Chapter 14 · Higher tier

GCSE Maths Quadratic and Simultaneous Equations (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

Factorise quadratic expressions, including the difference of two squares (and ax² + bx + c at Higher)
Solve quadratic equations by factorising, using the null factor law, after rearranging if necessary
Form and solve quadratic equations from areas, consecutive numbers and other problems
Solve simultaneous linear equations by elimination and by substitution
Solve simultaneous equations from graphs, and form them from words
Higher: use the quadratic formula, complete the square, solve linear/quadratic simultaneous equations and quadratic equations with fractions

You should already know

Expanding single and double brackets (Chapter 7)
Solving linear equations (Chapter 9)
Plotting straight-line and quadratic graphs (Chapters 12 and 13)
Square roots and surds (Chapter 6)

Where this leads

Quadratic inequalities (Chapter 10)
Equations of circles and tangents (Chapter 25)
Algebraic proof and rates of change in higher mathematics
Pythagoras and area problems (Chapters 18 and 20)

Edexcel GCSE (9–1) Mathematics 1MA1, Higher tier. Content references: A4, A18, A19, A21.

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

14.1 Factorising quadratic expressions: Worksheet (PDF) · Answers (PDF) · lesson page
14.2 Solving quadratic equations by factorising: Worksheet (PDF) · Answers (PDF) · lesson page
14.3 Rearranging first, and harder quadratics: Worksheet (PDF) · Answers (PDF) · lesson page
14.4 Problems leading to quadratic equations: Worksheet (PDF) · Answers (PDF) · lesson page
14.5 Simultaneous equations: elimination: Worksheet (PDF) · Answers (PDF) · lesson page
14.6 Simultaneous equations: substitution: Worksheet (PDF) · Answers (PDF) · lesson page
14.7 Solving equations with graphs: Worksheet (PDF) · Answers (PDF) · lesson page
14.8 Forming simultaneous equations from words: Worksheet (PDF) · Answers (PDF) · lesson page
14.9 The quadratic formula: Worksheet (PDF) · Answers (PDF) · lesson page
14.10 Completing the square: Worksheet (PDF) · Answers (PDF) · lesson page
14.11 Linear and quadratic simultaneous equations: Worksheet (PDF) · Answers (PDF) · lesson page
14.12 Quadratic equations with fractions: Worksheet (PDF) · Answers (PDF) · lesson page
Check-up and exam-style questions: Worksheet (PDF) · Answers (PDF) · lesson page

Exercise P Do you remember?

Non-calculator

FExpand and simplify (x + 3)(x + 5).

Done

FSolve 3x − 7 = 11.

Done

FFactorise 6x + 15.

Done

FWork out the value of x² − 3x when x = −2.

Done

FIs the point (2, 5) on the line y = 2x + 1? Show how you decide.

Done

14.1

Factorising quadratic expressions

Edexcel 1MA1: A4
Grade 3-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Factorising quadratic expressionsComing soon

In Chapter 7 you expanded brackets: (x + 2)(x + 3) = x² + 5x + 6. Factorising is the same journey backwards. You start with x² + 5x + 6 and look for the brackets that multiply to make it. You need this skill to solve quadratic equations, so it is worth becoming really fluent at it before you move on.

A quadratic expression is an expression whose highest power is 2, such as x² + 5x + 6. A factor is something that is multiplied: (x + 2) and (x + 3) are factors of x² + 5x + 6.

A rectangle with sides (x + 2) and (x + 3) is cut into four pieces. The two middle pieces, 3x and 2x, join to make 5x.

Look at the picture. The whole rectangle has area x² + 3x + 2x + 6 = x² + 5x + 6, and its sides are (x + 2) and (x + 3). Now look at the numbers: 2 × 3 = 6, which is the last number, and 2 + 3 = 5, which is the middle number (the number in front of x). This happens every time, and it gives us a method.

To factorise x² + bx + c: find two numbers that multiply to make c and add to make b. Call them p and q. Then the answer is (x + p)(x + q). The signs follow a pattern. If c is positive, both numbers have the same sign as b. If c is negative, one number is positive and one is negative, and the bigger one has the sign of b.

Always look for a common factor first. In 3x² − 12x every term divides by 3x, so it factorises as 3x(x − 4). Check by expanding.

Difference of two squares. An expression with two terms, both perfect squares, with a minus between them, factorises as a² − b² = (a − b)(a + b). Why? Expand (a − b)(a + b) and the two middle terms, +ab and −ab, cancel out. So x² − 49 = (x − 7)(x + 7). There is no middle term because there is no x term.

Quadratics where the number in front of x² is not 1 (like 2x² + 7x + 3) need one more step, called splitting the middle term. Multiply the first and last numbers (2 × 3 = 6). Find two numbers that multiply to make that (6) and add to make the middle number (7): they are 1 and 6. Split 7x into 1x + 6x, then factorise the front pair and the back pair separately: 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3). The matching brackets show it has worked.

Key ideas

Factorising is the reverse of expanding. Always check by expanding your answer.
x² + bx + c = (x + p)(x + q) where p × q = c and p + q = b.
Take out a common factor first: 3x² − 12x = 3x(x − 4).
Difference of two squares: a² − b² = (a − b)(a + b).
Higher: for ax² + bx + c, find two numbers that multiply to ac and add to b, split the middle term, and factorise in pairs.

Worked example 14.1.1

Factorise x² + 7x + 12.

▶ Watch this solution as a video

Key words. To factorise is the opposite of expanding: we turn a sum like x² + 5x + 6 into a product like (x + 2)(x + 3).

Method. Factorising means writing an expression as a product (brackets multiplied together).

Look at the expression and decide which method to use. Check in this order: (1) is there a common factor? (2) is it a difference of two squares (two square terms with a minus between)? (3) is it a quadratic with three terms?
x² + 7x + 12
Factorise. Follow the method for the pattern you found.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 7 and c = 12.
We need two numbers that multiply to 12 (the last number) and add to 7 (the middle number). List the pairs that multiply to 12:
1 × 12 = 12, add: 1 + 12 = 13 ✗
-1 × -12 = 12, add: -1 + -12 = -13 ✗
2 × 6 = 12, add: 2 + 6 = 8 ✗
-2 × -6 = 12, add: -2 + -6 = -8 ✗
3 × 4 = 12, add: 3 + 4 = 7 ✓ this pair works
Put the pair into two brackets: (x + 3)(x + 4).
Check by expanding: x² + 4x + 3x + 12 = x² + 7x + 12. ✓
x² + 7x + 12 = (x + 3)(x + 4)
Check by expanding. Factorising is the reverse of expanding, so multiplying the brackets out must give the original expression back.
First: x × x = x²
Outer: x × 4 = 4x
Inner: 3 × x = 3x
Last: 3 × 4 = 12
(x + 3)(x + 4) = x² + 7x + 12 ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong.

Worked example 14.1.2

Factorise x² − 2x − 15.

▶ Watch this solution as a video

Key words. To factorise is the opposite of expanding: we turn a sum like x² + 5x + 6 into a product like (x + 2)(x + 3).

Method. Factorising means writing an expression as a product (brackets multiplied together).

Look at the expression and decide which method to use. Check in this order: (1) is there a common factor? (2) is it a difference of two squares (two square terms with a minus between)? (3) is it a quadratic with three terms?
x² − 2x − 15
Factorise. Follow the method for the pattern you found.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -2 and c = -15.
We need two numbers that multiply to -15 (the last number) and add to -2 (the middle number). List the pairs that multiply to -15:
1 × -15 = -15, add: 1 + -15 = -14 ✗
-1 × 15 = -15, add: -1 + 15 = 14 ✗
3 × -5 = -15, add: 3 + -5 = -2 ✓ this pair works
Put the pair into two brackets: (x + 3)(x − 5).
Check by expanding: x² − 5x + 3x − 15 = x² − 2x − 15. ✓
x² − 2x − 15 = (x + 3)(x − 5)
Check by expanding. Factorising is the reverse of expanding, so multiplying the brackets out must give the original expression back.
First: x × x = x²
Outer: x × 3 = 3x
Inner: (−5) × x = −5x
Last: (−5) × 3 = −15
(x + 3)(x − 5) = x² − 2x − 15 ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong.

Worked example 14.1.3

Factorise x² − 49.

▶ Watch this solution as a video

Key words. To factorise is the opposite of expanding: we turn a sum like x² + 5x + 6 into a product like (x + 2)(x + 3).

Method. Factorising means writing an expression as a product (brackets multiplied together).

Look at the expression and decide which method to use. Check in this order: (1) is there a common factor? (2) is it a difference of two squares (two square terms with a minus between)? (3) is it a quadratic with three terms?
x² − 49
Factorise. Follow the method for the pattern you found.
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. x² = (x)², so a = x. And 49 = 7², so b = 7.
Put them into the rule: (x − 7)(x + 7).
Check by expanding: (x − 7)(x + 7) = x² + 7x − 7x − 49 = x² − 49. ✓ (the middle terms cancel)
x² − 49 = (x − 7)(x + 7)
Check by expanding. Factorising is the reverse of expanding, so multiplying the brackets out must give the original expression back.
First: x × x = x²
Outer: x × 7 = 7x
Inner: (−7) × x = −7x
Last: (−7) × 7 = −49
(x − 7)(x + 7) = x² − 49 ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong.

Worked example 14.1.4

Factorise fully 5x² − 20.

▶ Watch this solution as a video

Key words. To factorise is the opposite of expanding: we turn a sum like x² + 5x + 6 into a product like (x + 2)(x + 3).

Method. Factorising means writing an expression as a product (brackets multiplied together).

Look at the expression and decide which method to use. Check in this order: (1) is there a common factor? (2) is it a difference of two squares (two square terms with a minus between)? (3) is it a quadratic with three terms?
5x² − 20
Factorise. Follow the method for the pattern you found.
Look for a common factor. A common factor is something that divides into every term.
The terms are 5x², −20. The biggest number and the highest power of x that divide into all of them is 5.
Take it outside a bracket, and inside the bracket put each term divided by it:
5x² ÷ 5 = x²
−20 ÷ 5 = −4
So 5x² − 20 = 5(x² − 4).
Check by expanding: 5 × (x² − 4) = 5 × x², 5 × (−4) → 5x² − 20. ✓
Can the bracket be factorised again? Look at (x² − 4).
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. x² = (x)², so a = x. And 4 = 2², so b = 2.
Put them into the rule: (x − 2)(x + 2).
Check by expanding: (x − 2)(x + 2) = x² + 2x − 2x − 4 = x² − 4. ✓ (the middle terms cancel)
5x² − 20 = 5(x − 2)(x + 2)
Check by expanding. Factorising is the reverse of expanding, so multiplying the brackets out must give the original expression back.
First: 5x × x = 5x²
Outer: 5x × 2 = 10x
Inner: (−10) × x = −10x
Last: (−10) × 2 = −20
5(x − 2)(x + 2) = 5x² − 20 ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong.

Worked example 14.1.5

Factorise 2x² + 7x + 3.

Key words. To factorise is the opposite of expanding: we turn a sum like x² + 5x + 6 into a product like (x + 2)(x + 3).

Method. Factorising means writing an expression as a product (brackets multiplied together).

Look at the expression and decide which method to use. Check in this order: (1) is there a common factor? (2) is it a difference of two squares (two square terms with a minus between)? (3) is it a quadratic with three terms?
2x² + 7x + 3
Factorise. Follow the method for the pattern you found.
Spot the pattern. A quadratic where the number in front of x² is 2, not 1. We use the split the middle term method.
Multiply the first and last numbers: 2 × (3) = 6. We need two numbers that multiply to 6 and add to 7 (the middle number).
1 × 6 = 6, add: 7 ✓ this pair works
Split the middle term (7x) into 1x and 6x: 2x² + 1x + 6x + 3.
Factorise each pair separately. First pair: 2x² + x = x(2x + 1). Second pair: 6x + 3 = 3(2x + 1).
The two brackets (2x + 1) are the same. That is how we know it has worked. Write: x(2x + 1) + 3(2x + 1).
Take the common bracket out: (2x + 1) multiplied by (x + 3).
Result: (x + 3)(2x + 1).
Check by expanding: (x + 3)(2x + 1) = 2x² + 7x + 3. ✓
2x² + 7x + 3 = (x + 3)(2x + 1)
Check by expanding. Factorising is the reverse of expanding, so multiplying the brackets out must give the original expression back.
First: x × 2x = 2x²
Outer: x × 1 = x
Inner: 3 × 2x = 6x
Last: 3 × 1 = 3
(x + 3)(2x + 1) = 2x² + 7x + 3 ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong.

Worked example 14.1.6

Factorise 6x² − x − 2.

Key words. To factorise is the opposite of expanding: we turn a sum like x² + 5x + 6 into a product like (x + 2)(x + 3).

Method. Factorising means writing an expression as a product (brackets multiplied together).

Look at the expression and decide which method to use. Check in this order: (1) is there a common factor? (2) is it a difference of two squares (two square terms with a minus between)? (3) is it a quadratic with three terms?
6x² − x − 2
Factorise. Follow the method for the pattern you found.
Spot the pattern. A quadratic where the number in front of x² is 6, not 1. We use the split the middle term method.
Multiply the first and last numbers: 6 × (-2) = -12. We need two numbers that multiply to -12 and add to -1 (the middle number).
1 × -12 = -12, add: -11 ✗
-1 × 12 = -12, add: 11 ✗
2 × -6 = -12, add: -4 ✗
-2 × 6 = -12, add: 4 ✗
3 × -4 = -12, add: -1 ✓ this pair works
Split the middle term (-1x) into 3x and -4x: 6x² + 3x − 4x − 2.
Factorise each pair separately. First pair: 6x² + 3x = 3x(2x + 1). Second pair: −4x − 2 = −2(2x + 1).
The two brackets (2x + 1) are the same. That is how we know it has worked. Write: 3x(2x + 1) − 2(2x + 1).
Take the common bracket out: (2x + 1) multiplied by (3x − 2).
Result: (2x + 1)(3x − 2).
Check by expanding: (2x + 1)(3x − 2) = 6x² − x − 2. ✓
6x² − x − 2 = (2x + 1)(3x − 2)
Check by expanding. Factorising is the reverse of expanding, so multiplying the brackets out must give the original expression back.
First: 2x × 3x = 6x²
Outer: 2x × (−2) = −4x
Inner: 1 × 3x = 3x
Last: 1 × (−2) = −2
(2x + 1)(3x − 2) = 6x² − x − 2 ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong.

Watch out

The expression x² + 9 cannot be factorised, because there is a plus between the squares. Only a minus gives a difference of two squares. Also, do not forget to take out a common factor first: 2x² − 50 is 2(x − 5)(x + 5), not just 2(x² − 25).

14.2

Solving quadratic equations by factorising

Edexcel 1MA1: A18
Grade 4-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Solving quadratic equations by factorisingComing soon

A quadratic equation is an equation that contains an x² term (and no higher power), for example x² + 5x + 6 = 0. To solve it means to find the values of x that make it true. A linear equation such as 2x + 1 = 9 has one solution. A quadratic equation usually has two solutions. You can see why from its graph (Chapter 13): the curve crosses the x-axis in two places, and each crossing is a solution of the equation “expression = 0”.

The curve y = x² − 3x − 10 crosses the x-axis where x = −2 and x = 5. These are the two solutions of x² − 3x − 10 = 0.

So how do we find them without drawing a graph? The key is an idea called the null factor law. Think about multiplying two numbers together. If the answer is 0, one of the numbers must have been 0, because 7 × 3 = 21, 7 × 1 = 7, and only 7 × 0 = 0. This is also true when the two things are brackets. If (x − 5)(x + 2) = 0 then either (x − 5) = 0 or (x + 2) = 0. So x = 5 or x = −2.

Most quadratics are not written as brackets, so first we have to factorise (section 14.1). The method is:

Make sure the equation has 0 on one side (the null factor law only works for a product that equals 0).
Factorise the other side.
Set each bracket equal to 0.
Solve each of the two little equations.
Check both answers in the original equation.

Some special cases are worth knowing. If there is no number term, such as x² − 5x = 0, take out the common factor x: x(x − 5) = 0, so x = 0 or x = 5. If it is a difference of two squares, such as x² − 49 = 0, then (x − 7)(x + 7) = 0 gives x = 7 or x = −7, usually written x = ±7. And if both brackets are identical, such as (x − 3)² = 0, there is only one solution, x = 3. This is called a repeated root; the curve just touches the x-axis.

Key ideas

A quadratic equation has an x² term. It usually has two solutions (also called roots).
Null factor law: if A × B = 0 then A = 0 or B = 0.
Steps: (1) 0 on one side, (2) factorise, (3) each bracket = 0, (4) solve, (5) check.
Never divide by x. For x² − 5x = 0 factorise to x(x − 5) = 0 so that you do not lose the answer x = 0.
The solutions of “expression = 0” are where the graph crosses the x-axis.

Worked example 14.2.1

Solve (x + 3)(x − 5) = 0.

▶ Watch this solution as a video

Key words. A quadratic equation is an equation whose highest power of the letter is 2 (it has an x² term), like x² + 5x + 6 = 0. To solve it means to find every value of x that makes it true. These values are called the solutions or roots. To factorise means to write it as brackets multiplied together.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

Write down the equation. The quadratic is already written as brackets multiplied together and equal to 0.
(x + 3)(x − 5) = 0
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x + 3)(x − 5) = 0
So (x − 5) = 0 or (x + 3) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 5 = 0
Add 5 to both sides: x = 5

x + 3 = 0
Subtract 3 from both sides: x = −3
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −3 or x = 5

Check. When x = −3: ((−3) − 5)((−3) + 3) = (−8) × 0 = 0. ✓
When x = 5: (5 − 5)(5 + 3) = 0 × 8 = 0. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.2.2

Solve x² + 7x + 12 = 0.

▶ Watch this solution as a video

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
x² + 7x + 12 = 0
Check that one side is 0. We can only use the null factor law when one side of the equation is exactly 0.
The right-hand side is already 0, so there is nothing to move.
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 7 and c = 12.
We need two numbers that multiply to 12 (the last number) and add to 7 (the middle number). List the pairs that multiply to 12:
1 × 12 = 12, add: 1 + 12 = 13 ✗
-1 × -12 = 12, add: -1 + -12 = -13 ✗
2 × 6 = 12, add: 2 + 6 = 8 ✗
-2 × -6 = 12, add: -2 + -6 = -8 ✗
3 × 4 = 12, add: 3 + 4 = 7 ✓ this pair works
Put the pair into two brackets: (x + 3)(x + 4).
Check by expanding: x² + 4x + 3x + 12 = x² + 7x + 12. ✓
x² + 7x + 12 = (x + 3)(x + 4)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x + 3)(x + 4) = 0
So (x + 3) = 0 or (x + 4) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x + 3 = 0
Subtract 3 from both sides: x = −3

x + 4 = 0
Subtract 4 from both sides: x = −4
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −4 or x = −3

Check. When x = −4: left side = (−4)² + 7 × (−4) + 12 = 16 − 28 + 12 = 0. It equals 0. ✓
When x = −3: left side = (−3)² + 7 × (−3) + 12 = 9 − 21 + 12 = 0. It equals 0. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.2.3

Solve x² − 5x = 0.

▶ Watch this solution as a video

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
x² − 5x = 0
Check that one side is 0. We can only use the null factor law when one side of the equation is exactly 0.
The right-hand side is already 0, so there is nothing to move.
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Look for a common factor. A common factor is something that divides into every term.
The terms are x², −5x. The biggest number and the highest power of x that divide into all of them is x.
Take it outside a bracket, and inside the bracket put each term divided by it:
x² ÷ x = x
−5x ÷ x = −5
So x² − 5x = x(x − 5).
Check by expanding: x × (x − 5) = x × x, x × (−5) → x² − 5x. ✓
x² − 5x = x(x − 5)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
x(x − 5) = 0
So (x − 5) = 0 or x = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 5 = 0
Add 5 to both sides: x = 5

x = 0
x = 0 is a solution.
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = 0 or x = 5

Check. When x = 0: left side = 0² − 5 × 0 = 0 + 0 = 0. It equals 0. ✓
When x = 5: left side = 5² − 5 × 5 = 25 − 25 = 0. It equals 0. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.2.4

Solve x² − 81 = 0.

▶ Watch this solution as a video

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
x² − 81 = 0
Check that one side is 0. We can only use the null factor law when one side of the equation is exactly 0.
The right-hand side is already 0, so there is nothing to move.
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. x² = (x)², so a = x. And 81 = 9², so b = 9.
Put them into the rule: (x − 9)(x + 9).
Check by expanding: (x − 9)(x + 9) = x² + 9x − 9x − 81 = x² − 81. ✓ (the middle terms cancel)
x² − 81 = (x − 9)(x + 9)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 9)(x + 9) = 0
So (x − 9) = 0 or (x + 9) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 9 = 0
Add 9 to both sides: x = 9

x + 9 = 0
Subtract 9 from both sides: x = −9
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −9 or x = 9

Check. When x = −9: left side = (−9)² − 81 = 81 − 81 = 0. It equals 0. ✓
When x = 9: left side = 9² − 81 = 81 − 81 = 0. It equals 0. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.2.5

Solve x² − 6x + 9 = 0.

▶ Watch this solution as a video

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
x² − 6x + 9 = 0
Check that one side is 0. We can only use the null factor law when one side of the equation is exactly 0.
The right-hand side is already 0, so there is nothing to move.
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -6 and c = 9.
We need two numbers that multiply to 9 (the last number) and add to -6 (the middle number). List the pairs that multiply to 9:
1 × 9 = 9, add: 1 + 9 = 10 ✗
-1 × -9 = 9, add: -1 + -9 = -10 ✗
3 × 3 = 9, add: 3 + 3 = 6 ✗
-3 × -3 = 9, add: -3 + -3 = -6 ✓ this pair works
Put the pair into two brackets: (x − 3)(x − 3).
Check by expanding: x² − 3x − 3x + 9 = x² − 6x + 9. ✓
x² − 6x + 9 = (x − 3)²
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 3)² = 0
So (x − 3) = 0.
A bracket that appears twice gives the same answer twice, so there is only one solution from it (a repeated root).
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 3 = 0
Add 3 to both sides: x = 3
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = 3

Check. When x = 3: left side = 3² − 6 × 3 + 9 = 9 − 18 + 9 = 0. It equals 0. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.2.6

Solve (2x − 3)(x + 4) = 0.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

Write down the equation. The quadratic is already written as brackets multiplied together and equal to 0.
(2x − 3)(x + 4) = 0
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(2x − 3)(x + 4) = 0
So (x + 4) = 0 or (2x − 3) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x + 4 = 0
Subtract 4 from both sides: x = −4

2x − 3 = 0
Add 3 to both sides: 2x = 3
Divide both sides by 2: x = 3 ÷ 2 = 32
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −4 or x = 32

Check. When x = −4: ((−4) + 4)(2 × (−4) − 3) = 0 × (−11) = 0. ✓
When x = 32: (32 + 4)(2 × 32 − 3) = 112 × 0 = 0. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.2.7

Solve 3x² − 27 = 0.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
3x² − 27 = 0
Check that one side is 0. We can only use the null factor law when one side of the equation is exactly 0.
The right-hand side is already 0, so there is nothing to move.
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Look for a common factor. A common factor is something that divides into every term.
The terms are 3x², −27. The biggest number and the highest power of x that divide into all of them is 3.
Take it outside a bracket, and inside the bracket put each term divided by it:
3x² ÷ 3 = x²
−27 ÷ 3 = −9
So 3x² − 27 = 3(x² − 9).
Check by expanding: 3 × (x² − 9) = 3 × x², 3 × (−9) → 3x² − 27. ✓
Can the bracket be factorised again? Look at (x² − 9).
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. x² = (x)², so a = x. And 9 = 3², so b = 3.
Put them into the rule: (x − 3)(x + 3).
Check by expanding: (x − 3)(x + 3) = x² + 3x − 3x − 9 = x² − 9. ✓ (the middle terms cancel)
3x² − 27 = 3(x − 3)(x + 3)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
3(x − 3)(x + 3) = 0
So (x − 3) = 0 or (x + 3) = 0.
The number 3 in front can never be 0, so it gives no solution. Ignore it.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 3 = 0
Add 3 to both sides: x = 3

x + 3 = 0
Subtract 3 from both sides: x = −3
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −3 or x = 3

Check. When x = −3: left side = 3 × (−3)² − 27 = 27 − 27 = 0. It equals 0. ✓
When x = 3: left side = 3 × 3² − 27 = 27 − 27 = 0. It equals 0. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Watch out

When you solve (x − 5) = 0 the answer is x = +5; for (x + 2) = 0 it is x = −2. The sign in the solution is the opposite of the sign in the bracket. And never set a bracket equal to a number other than 0: if (x − 1)(x + 4) = 10, you cannot say x − 1 = 10. You must rearrange to equal 0 first (see the next section).

14.3

Rearranging first, and harder quadratics

Edexcel 1MA1: A18
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Rearranging first, and harder quadraticsComing soon

The null factor law only works when one side is exactly 0. Many quadratic equations do not start in that form, for example x² + 3x = 10. If you just factorised the left side you would get x(x + 3) = 10, and there is nothing special about a product equal to 10. So the first job is to rearrange the equation to make one side 0.

Use the same balance method as for linear equations (Chapter 9): do the same thing to both sides. To get x² + 3x = 10 into the right form, subtract 10 from both sides: x² + 3x − 10 = 0. Now factorise and carry on as before. A good habit is to move everything to the side where the x² term is positive, so that the number in front of x² is positive and the factorising is easier.

If the equation has brackets, expand them first, then collect like terms, then rearrange. For example (x + 2)(x + 5) = 28 becomes x² + 7x + 10 = 28, and then x² + 7x − 18 = 0.

Why is it wrong to say “x + 2 = 28 or x + 5 = 28”? Try it: if x + 2 = 28 then x = 26, and then x + 5 = 31, so the product is 28 × 31, which is nowhere near 28. Only a product equal to zero forces one of the factors to be zero.

Higher. When the number in front of x² is not 1, use the “split the middle term” method from section 14.1. For 3x² + 5x − 2 = 0, multiply 3 × (−2) = −6 and find two numbers that multiply to −6 and add to 5: they are 6 and −1. Split: 3x² + 6x − x − 2 = 3x(x + 2) − 1(x + 2) = (3x − 1)(x + 2). The solutions can be fractions: when 3x − 1 = 0, x = 1/3.

Key ideas

The null factor law needs 0 on one side. Rearrange first: do the same to both sides.
Expand brackets and collect like terms before you rearrange.
Move everything to the side where x² is positive, then factorise, set each bracket to 0 and solve.
Never set the brackets equal to a number other than 0.
Higher: for ax² + bx + c = 0, split the middle term. The solutions can be fractions.

Worked example 14.3.1

Solve x² + 3x = 10.

▶ Watch this solution as a video

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
x² + 3x = 10
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 10 from both sides: x² + 3x − 10 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 3 and c = -10.
We need two numbers that multiply to -10 (the last number) and add to 3 (the middle number). List the pairs that multiply to -10:
1 × -10 = -10, add: 1 + -10 = -9 ✗
-1 × 10 = -10, add: -1 + 10 = 9 ✗
2 × -5 = -10, add: 2 + -5 = -3 ✗
-2 × 5 = -10, add: -2 + 5 = 3 ✓ this pair works
Put the pair into two brackets: (x − 2)(x + 5).
Check by expanding: x² + 5x − 2x − 10 = x² + 3x − 10. ✓
x² + 3x − 10 = (x − 2)(x + 5)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 2)(x + 5) = 0
So (x − 2) = 0 or (x + 5) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 2 = 0
Add 2 to both sides: x = 2

x + 5 = 0
Subtract 5 from both sides: x = −5
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −5 or x = 2

Check. When x = −5: left side = (−5)² + 3 × (−5) = 25 − 15 = 10; right side = 10 = 10. Both sides equal. ✓
When x = 2: left side = 2² + 3 × 2 = 4 + 6 = 10; right side = 10 = 10. Both sides equal. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.3.2

Solve x² = 2x + 15.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
x² = 2x + 15
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 2x from both sides: x² − 2x = 15
Subtract 15 from both sides: x² − 2x − 15 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -2 and c = -15.
We need two numbers that multiply to -15 (the last number) and add to -2 (the middle number). List the pairs that multiply to -15:
1 × -15 = -15, add: 1 + -15 = -14 ✗
-1 × 15 = -15, add: -1 + 15 = 14 ✗
3 × -5 = -15, add: 3 + -5 = -2 ✓ this pair works
Put the pair into two brackets: (x + 3)(x − 5).
Check by expanding: x² − 5x + 3x − 15 = x² − 2x − 15. ✓
x² − 2x − 15 = (x + 3)(x − 5)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x + 3)(x − 5) = 0
So (x − 5) = 0 or (x + 3) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 5 = 0
Add 5 to both sides: x = 5

x + 3 = 0
Subtract 3 from both sides: x = −3
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −3 or x = 5

Check. When x = −3: left side = (−3)² = 9 = 9; right side = 2 × (−3) + 15 = 9. Both sides equal. ✓
When x = 5: left side = 5² = 25 = 25; right side = 2 × 5 + 15 = 25. Both sides equal. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.3.3

Solve (x + 2)(x + 5) = 28.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
(x + 2)(x + 5) = 28
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand (x + 2)(x + 5):
First: x × x = x²
Outer: x × 5 = 5x
Inner: 2 × x = 2x
Last: 2 × 5 = 10
so (x + 2)(x + 5) = x² + 7x + 10
The equation is now: x² + 7x + 10 = 28
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 28 from both sides: x² + 7x + 10 − 28 = 0
Collect like terms: x² + 7x − 18 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 7 and c = -18.
We need two numbers that multiply to -18 (the last number) and add to 7 (the middle number). List the pairs that multiply to -18:
1 × -18 = -18, add: 1 + -18 = -17 ✗
-1 × 18 = -18, add: -1 + 18 = 17 ✗
2 × -9 = -18, add: 2 + -9 = -7 ✗
-2 × 9 = -18, add: -2 + 9 = 7 ✓ this pair works
Put the pair into two brackets: (x − 2)(x + 9).
Check by expanding: x² + 9x − 2x − 18 = x² + 7x − 18. ✓
x² + 7x − 18 = (x − 2)(x + 9)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 2)(x + 9) = 0
So (x − 2) = 0 or (x + 9) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 2 = 0
Add 2 to both sides: x = 2

x + 9 = 0
Subtract 9 from both sides: x = −9
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −9 or x = 2

Check. When x = −9: left side = (−9)² + 7 × (−9) + 10 = 81 − 63 + 10 = 28; right side = 28 = 28. Both sides equal. ✓
When x = 2: left side = 2² + 7 × 2 + 10 = 4 + 14 + 10 = 28; right side = 28 = 28. Both sides equal. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.3.4

Solve x(x + 4) = 21.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
x(x + 4) = 21
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand x(x + 4):
x × x = x²
x × 4 = 4x
so x(x + 4) = x² + 4x
The equation is now: x² + 4x = 21
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 21 from both sides: x² + 4x − 21 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 4 and c = -21.
We need two numbers that multiply to -21 (the last number) and add to 4 (the middle number). List the pairs that multiply to -21:
1 × -21 = -21, add: 1 + -21 = -20 ✗
-1 × 21 = -21, add: -1 + 21 = 20 ✗
3 × -7 = -21, add: 3 + -7 = -4 ✗
-3 × 7 = -21, add: -3 + 7 = 4 ✓ this pair works
Put the pair into two brackets: (x − 3)(x + 7).
Check by expanding: x² + 7x − 3x − 21 = x² + 4x − 21. ✓
x² + 4x − 21 = (x − 3)(x + 7)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 3)(x + 7) = 0
So (x − 3) = 0 or (x + 7) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 3 = 0
Add 3 to both sides: x = 3

x + 7 = 0
Subtract 7 from both sides: x = −7
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −7 or x = 3

Check. When x = −7: left side = (−7)² + 4 × (−7) = 49 − 28 = 21; right side = 21 = 21. Both sides equal. ✓
When x = 3: left side = 3² + 4 × 3 = 9 + 12 = 21; right side = 21 = 21. Both sides equal. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.3.5

Solve 2x² + x = 15.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
2x² + x = 15
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 15 from both sides: 2x² + x − 15 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. A quadratic where the number in front of x² is 2, not 1. We use the split the middle term method.
Multiply the first and last numbers: 2 × (-15) = -30. We need two numbers that multiply to -30 and add to 1 (the middle number).
1 × -30 = -30, add: -29 ✗
-1 × 30 = -30, add: 29 ✗
2 × -15 = -30, add: -13 ✗
-2 × 15 = -30, add: 13 ✗
3 × -10 = -30, add: -7 ✗
-3 × 10 = -30, add: 7 ✗
5 × -6 = -30, add: -1 ✗
-5 × 6 = -30, add: 1 ✓ this pair works
Split the middle term (1x) into -5x and 6x: 2x² − 5x + 6x − 15.
Factorise each pair separately. First pair: 2x² − 5x = x(2x − 5). Second pair: 6x − 15 = 3(2x − 5).
The two brackets (2x − 5) are the same. That is how we know it has worked. Write: x(2x − 5) + 3(2x − 5).
Take the common bracket out: (2x − 5) multiplied by (x + 3).
Result: (x + 3)(2x − 5).
Check by expanding: (x + 3)(2x − 5) = 2x² + x − 15. ✓
2x² + x − 15 = (x + 3)(2x − 5)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x + 3)(2x − 5) = 0
So (x + 3) = 0 or (2x − 5) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x + 3 = 0
Subtract 3 from both sides: x = −3

2x − 5 = 0
Add 5 to both sides: 2x = 5
Divide both sides by 2: x = 5 ÷ 2 = 52
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −3 or x = 52

Check. When x = −3: left side = 2 × (−3)² + (−3) = 18 − 3 = 15; right side = 15 = 15. Both sides equal. ✓
When x = 52: left side = 2 × 52² + 52 = 252 + 52 = 15; right side = 15 = 15. Both sides equal. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.3.6

Solve 3x² + 5x = 2.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
3x² + 5x = 2
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 2 from both sides: 3x² + 5x − 2 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. A quadratic where the number in front of x² is 3, not 1. We use the split the middle term method.
Multiply the first and last numbers: 3 × (-2) = -6. We need two numbers that multiply to -6 and add to 5 (the middle number).
1 × -6 = -6, add: -5 ✗
-1 × 6 = -6, add: 5 ✓ this pair works
Split the middle term (5x) into -1x and 6x: 3x² − 1x + 6x − 2.
Factorise each pair separately. First pair: 3x² − x = x(3x − 1). Second pair: 6x − 2 = 2(3x − 1).
The two brackets (3x − 1) are the same. That is how we know it has worked. Write: x(3x − 1) + 2(3x − 1).
Take the common bracket out: (3x − 1) multiplied by (x + 2).
Result: (x + 2)(3x − 1).
Check by expanding: (x + 2)(3x − 1) = 3x² + 5x − 2. ✓
3x² + 5x − 2 = (x + 2)(3x − 1)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x + 2)(3x − 1) = 0
So (x + 2) = 0 or (3x − 1) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x + 2 = 0
Subtract 2 from both sides: x = −2

3x − 1 = 0
Add 1 to both sides: 3x = 1
Divide both sides by 3: x = 1 ÷ 3 = 13
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −2 or x = 13

Check. When x = −2: left side = 3 × (−2)² + 5 × (−2) = 12 − 10 = 2; right side = 2 = 2. Both sides equal. ✓
When x = 13: left side = 3 × 13² + 5 × 13 = 13 + 53 = 2; right side = 2 = 2. Both sides equal. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Watch out

Do not divide both sides by x to “simplify”. If x² = 9x you must not divide by x, because that loses the solution x = 0. Move everything to one side: x² − 9x = 0, then factorise x(x − 9) = 0, which gives x = 0 or x = 9. Also take care with signs when you move terms across the equals sign.

14.4

Problems leading to quadratic equations

Edexcel 1MA1: A18, A21
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Problems leading to quadratic equationsComing soon

Quadratic equations appear whenever two lengths are multiplied together, for example in areas, or whenever two numbers are multiplied together, for example “consecutive numbers”. Here is a strategy that works for every problem in this section.

Choose a letter for the unknown and write down what it stands for, with units.
Write every other quantity in terms of that letter (a rectangle that is 3 cm longer than it is wide has length x + 3 if the width is x).
Form an equation from the information in the question (for instance “area = 54”).
Expand, rearrange to equal 0, factorise and solve exactly as in the last two sections.
Decide which answers make sense. A length cannot be negative, so a negative solution is rejected. Then answer the question that was asked.
Check in the words of the problem, not just in the equation.

Consecutive numbers follow each other with no gaps: 8, 9, 10. If the first is n the next is n + 1. Consecutive odd numbers (or consecutive even numbers) are 2 apart: n and n + 2. The product of two consecutive numbers is n(n + 1).

A rectangle that is 3 cm longer than it is wide. If the area is 54 cm², then x(x + 3) = 54.

Higher. Problems can also use Pythagoras’ theorem. For a right-angled triangle with shorter sides a and b and hypotenuse c, a² + b² = c² (see Chapter 20). When the sides are given as expressions with x, you must expand squared brackets such as (x + 7)² = (x + 7)(x + 7) carefully. It does not equal x² + 49.

Key ideas

Choose a letter, write the other quantities using it, form an equation, solve, then interpret.
Reject answers that cannot happen (a negative length or a negative number of items).
Consecutive whole numbers: n, n + 1. Consecutive odd (or even) numbers: n, n + 2.
Area of a rectangle = length × width; area of a triangle = ½ × base × height.
Check your answer against the words of the question.

Worked example 14.4.1

A rectangle has width x cm and length (x + 3) cm. Its area is 54 cm². Find the width and the length.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the width be x cm. Then the length is (x + 3) cm.
Form an equation from the information. Area of a rectangle = length × width.
x(x + 3) = 54
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand x(x + 3):
x × x = x²
x × 3 = 3x
so x(x + 3) = x² + 3x
The equation is now: x² + 3x = 54
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 54 from both sides: x² + 3x − 54 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 3 and c = -54.
We need two numbers that multiply to -54 (the last number) and add to 3 (the middle number). List the pairs that multiply to -54:
1 × -54 = -54, add: 1 + -54 = -53 ✗
-1 × 54 = -54, add: -1 + 54 = 53 ✗
2 × -27 = -54, add: 2 + -27 = -25 ✗
-2 × 27 = -54, add: -2 + 27 = 25 ✗
3 × -18 = -54, add: 3 + -18 = -15 ✗
-3 × 18 = -54, add: -3 + 18 = 15 ✗
6 × -9 = -54, add: 6 + -9 = -3 ✗
-6 × 9 = -54, add: -6 + 9 = 3 ✓ this pair works
Put the pair into two brackets: (x − 6)(x + 9).
Check by expanding: x² + 9x − 6x − 54 = x² + 3x − 54. ✓
x² + 3x − 54 = (x − 6)(x + 9)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 6)(x + 9) = 0
So (x − 6) = 0 or (x + 9) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 6 = 0
Add 6 to both sides: x = 6

x + 9 = 0
Subtract 9 from both sides: x = −9
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −9 or x = 6
Decide which solution makes sense. x is a width, so it cannot be negative. Reject x = −9.
x = 6 cm
Find the length and answer the question. The length is x + 3.
Length = 6 + 3 = 9 cm
Width = 6 cm, length = 9 cm

Check. Check in the words: width 6 cm and length 9 cm differ by 3 cm and the area is 6 × 9 = 54 cm². ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Worked example 14.4.2

The product of two consecutive positive whole numbers is 156. Find the numbers.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the smaller number be n. The next whole number is n + 1.
Form an equation from the information. The product of the two numbers means they are multiplied.
n(n + 1) = 156
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand n(n + 1):
n × n = n²
n × 1 = n
so n(n + 1) = n² + n
The equation is now: n² + n = 156
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 156 from both sides: n² + n − 156 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form n² + bn + c, with b = 1 and c = -156.
We need two numbers that multiply to -156 (the last number) and add to 1 (the middle number). List the pairs that multiply to -156:
1 × -156 = -156, add: 1 + -156 = -155 ✗
-1 × 156 = -156, add: -1 + 156 = 155 ✗
2 × -78 = -156, add: 2 + -78 = -76 ✗
-2 × 78 = -156, add: -2 + 78 = 76 ✗
3 × -52 = -156, add: 3 + -52 = -49 ✗
-3 × 52 = -156, add: -3 + 52 = 49 ✗
4 × -39 = -156, add: 4 + -39 = -35 ✗
-4 × 39 = -156, add: -4 + 39 = 35 ✗
6 × -26 = -156, add: 6 + -26 = -20 ✗
-6 × 26 = -156, add: -6 + 26 = 20 ✗
12 × -13 = -156, add: 12 + -13 = -1 ✗
-12 × 13 = -156, add: -12 + 13 = 1 ✓ this pair works
Put the pair into two brackets: (n − 12)(n + 13).
Check by expanding: n² + 13n − 12n − 156 = n² + n − 156. ✓
n² + n − 156 = (n − 12)(n + 13)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(n − 12)(n + 13) = 0
So (n − 12) = 0 or (n + 13) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
n − 12 = 0
Add 12 to both sides: n = 12

n + 13 = 0
Subtract 13 from both sides: n = −13
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
n = −13 or n = 12
Decide which solution makes sense. The numbers are positive, so reject n = −13.
n = 12
Find both numbers. The numbers are n and n + 1.
12 and 12 + 1 = 13
The numbers are 12 and 13.

Check. Check: 12 × 13 = 156. ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Worked example 14.4.3

A triangle has base (x + 4) cm and height x cm. Its area is 30 cm². Find the height and the base.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the height of the triangle be x cm. Then the base is (x + 4) cm.
Form an equation from the information. The area of a triangle is ½ × base × height. Here ½ × (x + 4) × x = 30. Multiply both sides by 2 to remove the half, so (x + 4) × x = 60.
x(x + 4) = 60
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand x(x + 4):
x × x = x²
x × 4 = 4x
so x(x + 4) = x² + 4x
The equation is now: x² + 4x = 60
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 60 from both sides: x² + 4x − 60 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 4 and c = -60.
We need two numbers that multiply to -60 (the last number) and add to 4 (the middle number). List the pairs that multiply to -60:
1 × -60 = -60, add: 1 + -60 = -59 ✗
-1 × 60 = -60, add: -1 + 60 = 59 ✗
2 × -30 = -60, add: 2 + -30 = -28 ✗
-2 × 30 = -60, add: -2 + 30 = 28 ✗
3 × -20 = -60, add: 3 + -20 = -17 ✗
-3 × 20 = -60, add: -3 + 20 = 17 ✗
4 × -15 = -60, add: 4 + -15 = -11 ✗
-4 × 15 = -60, add: -4 + 15 = 11 ✗
5 × -12 = -60, add: 5 + -12 = -7 ✗
-5 × 12 = -60, add: -5 + 12 = 7 ✗
6 × -10 = -60, add: 6 + -10 = -4 ✗
-6 × 10 = -60, add: -6 + 10 = 4 ✓ this pair works
Put the pair into two brackets: (x − 6)(x + 10).
Check by expanding: x² + 10x − 6x − 60 = x² + 4x − 60. ✓
x² + 4x − 60 = (x − 6)(x + 10)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 6)(x + 10) = 0
So (x − 6) = 0 or (x + 10) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 6 = 0
Add 6 to both sides: x = 6

x + 10 = 0
Subtract 10 from both sides: x = −10
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −10 or x = 6
Decide which solution makes sense. A height cannot be negative, so reject x = −10.
x = 6 cm
Find the base and answer. The base is x + 4.
Base = 10 cm
Height = 6 cm, base = 10 cm

Check. Area = ½ × 10 × 6 = 30 cm². ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Worked example 14.4.4

The product of two consecutive positive odd numbers is 143. Find the numbers.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the smaller number be n. The next consecutive odd number is n + 2 (odd numbers are 2 apart).
Form an equation from the information. The product of the two numbers means they are multiplied.
n(n + 2) = 143
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand n(n + 2):
n × n = n²
n × 2 = 2n
so n(n + 2) = n² + 2n
The equation is now: n² + 2n = 143
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 143 from both sides: n² + 2n − 143 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form n² + bn + c, with b = 2 and c = -143.
We need two numbers that multiply to -143 (the last number) and add to 2 (the middle number). List the pairs that multiply to -143:
1 × -143 = -143, add: 1 + -143 = -142 ✗
-1 × 143 = -143, add: -1 + 143 = 142 ✗
11 × -13 = -143, add: 11 + -13 = -2 ✗
-11 × 13 = -143, add: -11 + 13 = 2 ✓ this pair works
Put the pair into two brackets: (n − 11)(n + 13).
Check by expanding: n² + 13n − 11n − 143 = n² + 2n − 143. ✓
n² + 2n − 143 = (n − 11)(n + 13)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(n − 11)(n + 13) = 0
So (n − 11) = 0 or (n + 13) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
n − 11 = 0
Add 11 to both sides: n = 11

n + 13 = 0
Subtract 13 from both sides: n = −13
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
n = −13 or n = 11
Decide which solution makes sense. The numbers are positive, so reject n = −13.
n = 11
Find both numbers. The numbers are n and n + 2.
11 and 11 + 2 = 13
The numbers are 11 and 13.

Check. Check: 11 × 13 = 143. ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Worked example 14.4.5

The sides of a right-angled triangle are x cm, (x + 1) cm and (x + 2) cm, where (x + 2) cm is the longest side. Find x.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the shortest side be x cm. The other two sides are (x + 1) cm and the hypotenuse (x + 2) cm.
Form an equation from the information. Pythagoras: (shorter side)² + (other side)² = (hypotenuse)². The longest side, the hypotenuse, is (x + 2).
x² + (x + 1)² = (x + 2)²
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand (x + 1)(x + 1):
First: x × x = x²
Outer: x × 1 = x
Inner: 1 × x = x
Last: 1 × 1 = 1
so (x + 1)(x + 1) = x² + 2x + 1

Expand (x + 2)(x + 2):
First: x × x = x²
Outer: x × 2 = 2x
Inner: 2 × x = 2x
Last: 2 × 2 = 4
so (x + 2)(x + 2) = x² + 4x + 4
The equation is now: 2x² + 2x + 1 = x² + 4x + 4
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract x² from both sides: 2x² + 2x + 1 − x² = 4x + 4
Subtract 4x from both sides: 2x² + 2x + 1 − x² − 4x = 4
Subtract 4 from both sides: 2x² + 2x + 1 − x² − 4x − 4 = 0
Collect like terms: x² − 2x − 3 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -2 and c = -3.
We need two numbers that multiply to -3 (the last number) and add to -2 (the middle number). List the pairs that multiply to -3:
1 × -3 = -3, add: 1 + -3 = -2 ✓ this pair works
Put the pair into two brackets: (x + 1)(x − 3).
Check by expanding: x² − 3x + 1x − 3 = x² − 2x − 3. ✓
x² − 2x − 3 = (x + 1)(x − 3)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x + 1)(x − 3) = 0
So (x − 3) = 0 or (x + 1) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 3 = 0
Add 3 to both sides: x = 3

x + 1 = 0
Subtract 1 from both sides: x = −1
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −1 or x = 3
Decide which solution makes sense. A length cannot be negative, so reject x = −1.
x = 3 cm
Find all three sides. Put x = 3 into each expression.
Sides: 3 cm, 3 + 1 = 4 cm and 3 + 2 = 5 cm
The sides are 3 cm, 4 cm and 5 cm.

Check. 3² + 4² = 9 + 16 = 25 = 5². ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Watch out

Always reject impossible answers, and always answer the question that was asked. If x is the width, the question may want the length or the area, so do not stop at x. Also remember (x + 1)² is not x² + 1: it is x² + 2x + 1.

14.5

Simultaneous equations: elimination

Edexcel 1MA1: A19
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Simultaneous equations: eliminationComing soon

One equation with two letters, such as x + y = 7, has lots of solutions: (1, 6), (2, 5), (3, 4), and so on. But if a second equation must be true at the same time, such as x − y = 1, there is usually only one pair of values that works for both: x = 4 and y = 3. Two equations that must both be true at the same time are called simultaneous equations. (“Simultaneous” means “at the same time”.)

Simultaneous equations from prices: 2p + b = 5 and p + b = 3. The first basket has one more pen, and costs £2 more, so a pen costs £2.

The picture shows the idea of elimination. The two baskets are the same except that the first has one extra pen, and it costs £2 more. So one pen costs £2, and then the book costs £1. We got rid of the book by comparing the baskets: that is elimination. In algebra we do this by adding or subtracting the two equations. This is allowed because if you add equal amounts to equal amounts, the results are equal.

The method.

Number the equations (1) and (2). Make sure each has the form (number)x + (number)y = (number).
Choose a letter to eliminate. Make the numbers in front of it the same size in both equations. If they are not, multiply one or both equations (multiply every term, including the right-hand side).
If the matching terms have the same sign, subtract the equations. If they have different signs, add them. One letter disappears.
Solve the equation that is left.
Substitute the value into one of the original equations to find the other letter.
Check in the other original equation.

A memory aid: Same Signs Subtract. If the equations are not lined up, rearrange them first: 2x = 13 − y becomes 2x + y = 13.

Key ideas

Simultaneous equations: two equations, two letters, both true at once. The solution is a pair of values.
Eliminate a letter by adding or subtracting the equations. Same signs: subtract. Different signs: add.
If no letter has matching numbers, multiply one or both equations first. Multiply every term.
Find one letter, substitute to find the other, and always check in the other equation.
Line up the equations first: letters on the left, numbers on the right.

Worked example 14.5.1

Solve the simultaneous equations 3x + 2y = 12 and 3x − y = 3.

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Simultaneous equations: make the number in front of one letter the same size in both equations, then add or subtract to eliminate (remove) that letter. Solve for the other letter, substitute back, and check.

Number the equations. We will keep referring to them, so call them (1) and (2).
(1) 3x + 2y = 12
(2) 3x − y = 3
Choose a letter to eliminate: x. To eliminate a letter, the numbers in front of it must be the same size in both equations. The numbers in front of x are 3 and 3: they are already the same size, so no multiplying is needed.
Subtract (2) from (1). The x terms are 3x and 3x. They have the same sign, so subtract the equations: 3x − 3x = 0, so x disappears. Do the same thing to the left sides and to the right sides.
(1) 3x + 2y = 12
(2) 3x − y = 3
x terms: 3x − 3x = 0 (they cancel)
y terms: 2y − (−y) = 3y
Numbers: 12 − 3 = 9
3y = 9
Solve for y. 3y means 3 multiplied by y, so divide both sides by 3.
Divide both sides by 3: y = 9 ÷ 3 = 3
Substitute y = 3 into equation (1) to find x. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
Equation (1): 3x + 2y = 12
3x + 2 × 3 = 12
3x + 6 = 12
Subtract 6 from both sides: 3x = 6
Divide both sides by 3: x = 6 ÷ 3 = 2
Write the solution. Give both values. They belong together as a pair.
x = 2, y = 3

Check. Put x = 2, y = 3 into the other equation (equation (2), which we did not use to find x): 3 × 2 − 3 = 6 − 3 = 3. This equals 3, as it should. ✓

Watch out. Same signs subtract, different signs add. Multiply every term of the equation, including the number on the right, and take care with negative numbers when you subtract.

Worked example 14.5.2

Solve 2x + 3y = 12 and 4x − 3y = 6.

Number the equations. We will keep referring to them, so call them (1) and (2).
(1) 2x + 3y = 12
(2) 4x − 3y = 6
Choose a letter to eliminate: y. To eliminate a letter, the numbers in front of it must be the same size in both equations. The numbers in front of y are 3 and −3: they are already the same size, so no multiplying is needed.
Add (1) and (2). The y terms are 3y and −3y. They have different signs, so add the equations: 3y + (−3y) = 0, so y disappears. Do the same thing to the left sides and to the right sides.
(1) 2x + 3y = 12
(2) 4x − 3y = 6
y terms: 3y + (−3y) = 0 (they cancel)
x terms: 2x + 4x = 6x
Numbers: 12 + 6 = 18
6x = 18
Solve for x. 6x means 6 multiplied by x, so divide both sides by 6.
Divide both sides by 6: x = 18 ÷ 6 = 3
Substitute x = 3 into equation (1) to find y. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
Equation (1): 2x + 3y = 12
2 × 3 + 3y = 12
3y + 6 = 12
Subtract 6 from both sides: 3y = 6
Divide both sides by 3: y = 6 ÷ 3 = 2
Write the solution. Give both values. They belong together as a pair.
x = 3, y = 2

Check. Put x = 3, y = 2 into the other equation (equation (2), which we did not use to find y): 4 × 3 − 3 × 2 = 12 − 6 = 6. This equals 6, as it should. ✓

Watch out. Same signs subtract, different signs add. Multiply every term of the equation, including the number on the right, and take care with negative numbers when you subtract.

Worked example 14.5.3

Solve 2x + y = 7 and 3x − 2y = 7. (Multiply one equation first.)

Number the equations. We will keep referring to them, so call them (1) and (2).
(1) 2x + y = 7
(2) 3x − 2y = 7
Choose a letter to eliminate: y, and make the numbers match. The numbers in front of y are 1 and 2. The lowest common multiple of 1 and 2 is 2, so we will make both equal to 2. So multiply equation (1) by 2. Multiply every term, including the number on the right-hand side.
Multiply to make the numbers in front of y match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
Equation (1) × 2: 2 × 2x, 2 × y on the left and 2 × 7 on the right.
(3) 4x + 2y = 14
Add (3) and (2). The y terms are 2y and −2y. They have different signs, so add the equations: 2y + (−2y) = 0, so y disappears. Do the same thing to the left sides and to the right sides.
(3) 4x + 2y = 14
(2) 3x − 2y = 7
y terms: 2y + (−2y) = 0 (they cancel)
x terms: 4x + 3x = 7x
Numbers: 14 + 7 = 21
7x = 21
Solve for x. 7x means 7 multiplied by x, so divide both sides by 7.
Divide both sides by 7: x = 21 ÷ 7 = 3
Substitute x = 3 into equation (1) to find y. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
Equation (1): 2x + y = 7
2 × 3 + y = 7
y + 6 = 7
Subtract 6 from both sides: y = 1
Write the solution. Give both values. They belong together as a pair.
x = 3, y = 1

Check. Put x = 3, y = 1 into the other equation (equation (2), which we did not use to find y): 3 × 3 − 2 × 1 = 9 − 2 = 7. This equals 7, as it should. ✓

Watch out. Same signs subtract, different signs add. Multiply every term of the equation, including the number on the right, and take care with negative numbers when you subtract.

Worked example 14.5.4

Solve 3x + 2y = 16 and 5x − 3y = 14. (Multiply both equations.)

Number the equations. We will keep referring to them, so call them (1) and (2).
(1) 3x + 2y = 16
(2) 5x − 3y = 14
Choose a letter to eliminate: y, and make the numbers match. The numbers in front of y are 2 and 3. The lowest common multiple of 2 and 3 is 6, so we will make both equal to 6. So multiply equation (1) by 3 and multiply equation (2) by 2. Multiply every term, including the number on the right-hand side.
Multiply to make the numbers in front of y match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
Equation (1) × 3: 3 × 3x, 3 × 2y on the left and 3 × 16 on the right.
(3) 9x + 6y = 48

Equation (2) × 2: 2 × 5x, 2 × (−3y) on the left and 2 × 14 on the right.
(4) 10x − 6y = 28
Add (3) and (4). The y terms are 6y and −6y. They have different signs, so add the equations: 6y + (−6y) = 0, so y disappears. Do the same thing to the left sides and to the right sides.
(3) 9x + 6y = 48
(4) 10x − 6y = 28
y terms: 6y + (−6y) = 0 (they cancel)
x terms: 9x + 10x = 19x
Numbers: 48 + 28 = 76
19x = 76
Solve for x. 19x means 19 multiplied by x, so divide both sides by 19.
Divide both sides by 19: x = 76 ÷ 19 = 4
Substitute x = 4 into equation (1) to find y. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
Equation (1): 3x + 2y = 16
3 × 4 + 2y = 16
2y + 12 = 16
Subtract 12 from both sides: 2y = 4
Divide both sides by 2: y = 4 ÷ 2 = 2
Write the solution. Give both values. They belong together as a pair.
x = 4, y = 2

Check. Put x = 4, y = 2 into the other equation (equation (2), which we did not use to find y): 5 × 4 − 3 × 2 = 20 − 6 = 14. This equals 14, as it should. ✓

Watch out. Same signs subtract, different signs add. Multiply every term of the equation, including the number on the right, and take care with negative numbers when you subtract.

Worked example 14.5.5

Solve 2x = 13 − y and x + 3y = 14. (Rearrange first.)

Rearrange each equation so the letters are on the left and the plain number is on the right. Elimination needs the equations lined up in the same pattern: (number)x + (number)y = (number). Use inverse operations on both sides.
2x = 13 − y → 2x + y = 13
x + 3y = 14 → x + 3y = 14
Number the equations. Call the tidied equations (1) and (2).
(1) 2x + y = 13
(2) x + 3y = 14
Choose a letter to eliminate: x, and make the numbers match. The numbers in front of x are 2 and 1. The lowest common multiple of 2 and 1 is 2, so we will make both equal to 2. So multiply equation (2) by 2. Multiply every term, including the number on the right-hand side.
Multiply to make the numbers in front of x match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
Equation (2) × 2: 2 × x, 2 × 3y on the left and 2 × 14 on the right.
(3) 2x + 6y = 28
Subtract (1) from (3). The x terms are 2x and 2x. They have the same sign, so subtract the equations: 2x − 2x = 0, so x disappears. Do the same thing to the left sides and to the right sides.
(3) 2x + 6y = 28
(1) 2x + y = 13
x terms: 2x − 2x = 0 (they cancel)
y terms: 6y − y = 5y
Numbers: 28 − 13 = 15
5y = 15
Solve for y. 5y means 5 multiplied by y, so divide both sides by 5.
Divide both sides by 5: y = 15 ÷ 5 = 3
Substitute y = 3 into equation (2) to find x. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
Equation (2): x + 3y = 14
x + 3 × 3 = 14
x + 9 = 14
Subtract 9 from both sides: x = 5
Write the solution. Give both values. They belong together as a pair.
x = 5, y = 3

Check. Put x = 5, y = 3 into the other equation (equation (1), which we did not use to find x): 2 × 5 + 3 = 10 + 3 = 13. This equals 13, as it should. ✓

Watch out. Same signs subtract, different signs add. Multiply every term of the equation, including the number on the right, and take care with negative numbers when you subtract.

Worked example 14.5.6

Solve 2x + 4y = 5 and 6x − 4y = 3. (The answer includes a fraction.)

Number the equations. We will keep referring to them, so call them (1) and (2).
(1) 2x + 4y = 5
(2) 6x − 4y = 3
Choose a letter to eliminate: y. To eliminate a letter, the numbers in front of it must be the same size in both equations. The numbers in front of y are 4 and −4: they are already the same size, so no multiplying is needed.
Add (1) and (2). The y terms are 4y and −4y. They have different signs, so add the equations: 4y + (−4y) = 0, so y disappears. Do the same thing to the left sides and to the right sides.
(1) 2x + 4y = 5
(2) 6x − 4y = 3
y terms: 4y + (−4y) = 0 (they cancel)
x terms: 2x + 6x = 8x
Numbers: 5 + 3 = 8
8x = 8
Solve for x. 8x means 8 multiplied by x, so divide both sides by 8.
Divide both sides by 8: x = 8 ÷ 8 = 1
Substitute x = 1 into equation (1) to find y. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
Equation (1): 2x + 4y = 5
2 × 1 + 4y = 5
4y + 2 = 5
Subtract 2 from both sides: 4y = 3
Divide both sides by 4: y = 3 ÷ 4 = 34
Write the solution. Give both values. They belong together as a pair.
x = 1, y = 34

Check. Put x = 1, y = 34 into the other equation (equation (2), which we did not use to find y): 6 × 1 − 4 × 34 = 6 − 3 = 3. This equals 3, as it should. ✓

Watch out. Same signs subtract, different signs add. Multiply every term of the equation, including the number on the right, and take care with negative numbers when you subtract.

Worked example 14.5.7

Solve 3x + 2y = 1 and 2x + 3y = −1. (Negative answers.)

Number the equations. We will keep referring to them, so call them (1) and (2).
(1) 3x + 2y = 1
(2) 2x + 3y = −1
Choose a letter to eliminate: x, and make the numbers match. The numbers in front of x are 3 and 2. The lowest common multiple of 3 and 2 is 6, so we will make both equal to 6. So multiply equation (1) by 2 and multiply equation (2) by 3. Multiply every term, including the number on the right-hand side.
Multiply to make the numbers in front of x match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
Equation (1) × 2: 2 × 3x, 2 × 2y on the left and 2 × 1 on the right.
(3) 6x + 4y = 2

Equation (2) × 3: 3 × 2x, 3 × 3y on the left and 3 × (−1) on the right.
(4) 6x + 9y = −3
Subtract (3) from (4). The x terms are 6x and 6x. They have the same sign, so subtract the equations: 6x − 6x = 0, so x disappears. Do the same thing to the left sides and to the right sides.
(4) 6x + 9y = −3
(3) 6x + 4y = 2
x terms: 6x − 6x = 0 (they cancel)
y terms: 9y − 4y = 5y
Numbers: −3 − 2 = −5
5y = −5
Solve for y. 5y means 5 multiplied by y, so divide both sides by 5.
Divide both sides by 5: y = −5 ÷ 5 = −1
Substitute y = −1 into equation (2) to find x. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
Equation (2): 2x + 3y = −1
2x + 3 × (−1) = −1
2x − 3 = −1
Add 3 to both sides: 2x = 2
Divide both sides by 2: x = 2 ÷ 2 = 1
Write the solution. Give both values. They belong together as a pair.
x = 1, y = −1

Check. Put x = 1, y = −1 into the other equation (equation (1), which we did not use to find x): 3 × 1 + 2 × (−1) = 3 − 2 = 1. This equals 1, as it should. ✓

Watch out. Same signs subtract, different signs add. Multiply every term of the equation, including the number on the right, and take care with negative numbers when you subtract.

Watch out

When you subtract, you subtract every term, and subtracting a negative number is the same as adding. For example (2y) − (−y) = 3y, not y. A very common mistake is to subtract only the first term, or to forget to multiply the number on the right-hand side when you multiply an equation.

14.6

Simultaneous equations: substitution

Edexcel 1MA1: A19
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Simultaneous equations: substitutionComing soon

Elimination is not the only way to solve simultaneous equations. Sometimes one equation already tells you what a letter equals, for example y = 2x + 1. Then it is quick to substitute, which means “replace”: wherever you see y in the other equation, write 2x + 1 instead. Because y and 2x + 1 are equal, this does not change the equation, but it removes y, leaving an ordinary equation with one letter.

The method.

Number the equations. Choose one that has a letter on its own (y = … or x = …). If there is not one, rearrange one equation to make a letter the subject. The easiest choice is a letter with no number in front of it.
Substitute this expression into the other equation, in brackets.
Expand the brackets and collect like terms.
Solve the one-letter equation.
Substitute the answer back into the rearranged equation to find the other letter.
Check in the other equation.

Brackets are very important. If y = 2x + 1 and the other equation is 4x − 3y = 7, then you must write 4x − 3(2x + 1) = 7. Without the brackets, the −3 would only multiply the 2x and you would get the wrong answer.

Which method should I use? If one equation has a letter on its own (or with a number 1 in front of it), substitution is usually quickest. If both equations are in the form ax + by = c and the numbers are awkward, elimination is usually quicker. Either method always gives the same answer, so you can use one to check the other.

Key ideas

Substitution: replace a letter by an expression that equals it.
Make a letter the subject of one equation (choose a letter with 1 in front of it).
Put the expression in brackets when you substitute, then expand and collect like terms.
Solve for one letter, then find the other, then check in the other equation.
Use whichever method (substitution or elimination) is easier. They give the same answer.

Worked example 14.6.1

Solve y = 2x + 1 and 3x + y = 11.

Method. Substitution: make one letter the subject of one equation, substitute it into the other equation (in brackets), solve for the remaining letter, then find the first letter.

Write the equations. We have two equations with two unknown letters. We will call them (1) and (2).
(1) y = 2x + 1
(2) 3x + y = 11
Equation (1) already tells us what y equals. y is on its own on the left, so we can use it as it is.
y = 2x + 1
Substitute into equation (2). Equation (2) contains y. Wherever y appears, write what it equals (inside a bracket). Now there is only one letter, x, so we can solve.
Equation (2): 3x + y = 11
3x + (2x + 1) = 11
Expand the bracket and collect like terms. Multiply the bracket out, then add the x terms together and the plain numbers together.
There is no number in front of the bracket, so the brackets can just be removed: (2x + 1) = 2x + 1
The equation becomes: 2x + 1 + 3x = 11
Collect like terms: 5x + 1 = 11
Solve the equation for x. It is now a one-letter equation. Use inverse operations on both sides.
Subtract 1 from both sides: 5x = 10
Divide both sides by 5: x = 10 ÷ 5 = 2
Substitute x = 2 into the rearranged equation to find y. The rearranged equation y = 2x + 1 is the easiest to use.
y = 2x + 1
y = 2 × 2 + 1
y = 5
Write the solution. Give both values as a pair.
x = 2, y = 5

Check. Put x = 2, y = 5 into equation (1): −2 × 2 + 5 = −4 + 5 = 1. It equals 1, as it should. ✓

Watch out. Put the expression in brackets when you substitute, and multiply the bracket by everything in front of it. A minus sign in front of a bracket changes every sign inside.

Worked example 14.6.2

Solve x = 3y − 2 and 2x + y = 10.

Method. Substitution: make one letter the subject of one equation, substitute it into the other equation (in brackets), solve for the remaining letter, then find the first letter.

Write the equations. We have two equations with two unknown letters. We will call them (1) and (2).
(1) x = 3y − 2
(2) 2x + y = 10
Rearrange equation (2) to make y the subject. The subject is the letter on its own on one side. We choose the equation and letter that is easiest to rearrange (a letter with no number in front of it, so there is no dividing).
Equation (2): 2x + y = 10
Subtract 2x from both sides: y = 10 − 2x
Substitute into equation (1). Equation (1) contains y. Wherever y appears, write what it equals (inside a bracket). Now there is only one letter, x, so we can solve.
Equation (1): x − 3y = −2
x − 3(10 − 2x) = −2
Expand the bracket and collect like terms. Multiply the bracket out, then add the x terms together and the plain numbers together.
Expand −3(10 − 2x):
(−3) × (−2x) = 6x
(−3) × 10 = −30
so −3(10 − 2x) = 6x − 30
The equation becomes: 6x − 30 + x = −2
Collect like terms: 7x − 30 = −2
Solve the equation for x. It is now a one-letter equation. Use inverse operations on both sides.
Add 30 to both sides: 7x = 28
Divide both sides by 7: x = 28 ÷ 7 = 4
Substitute x = 4 into the rearranged equation to find y. The rearranged equation y = 10 − 2x is the easiest to use.
y = 10 − 2x
y = 10 − 2 × 4
y = 2
Write the solution. Give both values as a pair.
x = 4, y = 2

Check. Put x = 4, y = 2 into equation (2): 2 × 4 + 2 = 8 + 2 = 10. It equals 10, as it should. ✓

Watch out. Put the expression in brackets when you substitute, and multiply the bracket by everything in front of it. A minus sign in front of a bracket changes every sign inside.

Worked example 14.6.3

Solve x + 2y = 8 and 3x − y = 3. (Rearrange one equation first.)

Method. Substitution: make one letter the subject of one equation, substitute it into the other equation (in brackets), solve for the remaining letter, then find the first letter.

Write the equations and number them. We have two unknown letters, so we need two equations.
(1) x + 2y = 8
(2) 3x − y = 3
Rearrange equation (1) to make x the subject. The subject is the letter on its own on one side. We choose the equation and letter that is easiest to rearrange (a letter with no number in front of it, so there is no dividing).
Equation (1): x + 2y = 8
Subtract 2y from both sides: x = 8 − 2y
Substitute into equation (2). Equation (2) contains x. Wherever x appears, write what it equals (inside a bracket). Now there is only one letter, y, so we can solve.
Equation (2): 3x − y = 3
3(8 − 2y) − y = 3
Expand the bracket and collect like terms. Multiply the bracket out, then add the y terms together and the plain numbers together.
Expand 3(8 − 2y):
3 × (−2y) = −6y
3 × 8 = 24
so 3(8 − 2y) = −6y + 24
The equation becomes: −6y + 24 − y = 3
Collect like terms: −7y + 24 = 3
Solve the equation for y. It is now a one-letter equation. Use inverse operations on both sides.
Subtract 24 from both sides: −7y = −21
Divide both sides by −7: y = −21 ÷ (−7) = 3
Substitute y = 3 into the rearranged equation to find x. The rearranged equation x = 8 − 2y is the easiest to use.
x = 8 − 2y
x = 8 − 2 × 3
x = 2
Write the solution. Give both values as a pair.
x = 2, y = 3

Check. Put x = 2, y = 3 into equation (1): 2 + 2 × 3 = 2 + 6 = 8. It equals 8, as it should. ✓

Watch out. Put the expression in brackets when you substitute, and multiply the bracket by everything in front of it. A minus sign in front of a bracket changes every sign inside.

Worked example 14.6.4

Solve y = x − 5 and 3x + 2y = 5. (The answer has a negative number.)

Method. Substitution: make one letter the subject of one equation, substitute it into the other equation (in brackets), solve for the remaining letter, then find the first letter.

Write the equations. We have two equations with two unknown letters. We will call them (1) and (2).
(1) y = x − 5
(2) 3x + 2y = 5
Equation (1) already tells us what y equals. y is on its own on the left, so we can use it as it is.
y = x − 5
Substitute into equation (2). Equation (2) contains y. Wherever y appears, write what it equals (inside a bracket). Now there is only one letter, x, so we can solve.
Equation (2): 3x + 2y = 5
3x + 2(x − 5) = 5
Expand the bracket and collect like terms. Multiply the bracket out, then add the x terms together and the plain numbers together.
Expand 2(x − 5):
2 × x = 2x
2 × (−5) = −10
so 2(x − 5) = 2x − 10
The equation becomes: 2x − 10 + 3x = 5
Collect like terms: 5x − 10 = 5
Solve the equation for x. It is now a one-letter equation. Use inverse operations on both sides.
Add 10 to both sides: 5x = 15
Divide both sides by 5: x = 15 ÷ 5 = 3
Substitute x = 3 into the rearranged equation to find y. The rearranged equation y = x − 5 is the easiest to use.
y = x − 5
y = 3 − 5
y = −2
Write the solution. Give both values as a pair.
x = 3, y = −2

Check. Put x = 3, y = −2 into equation (1): −3 + (−2) = −3 − 2 = −5. It equals −5, as it should. ✓

Watch out. Put the expression in brackets when you substitute, and multiply the bracket by everything in front of it. A minus sign in front of a bracket changes every sign inside.

Worked example 14.6.5

Solve y = 7 − 2x and 3x − 2y = 7. (Take care with the minus sign in front of a bracket.)

Method. Substitution: make one letter the subject of one equation, substitute it into the other equation (in brackets), solve for the remaining letter, then find the first letter.

Write the equations. We have two equations with two unknown letters. We will call them (1) and (2).
(1) y = 7 − 2x
(2) 3x − 2y = 7
Equation (1) already tells us what y equals. y is on its own on the left, so we can use it as it is.
y = 7 − 2x
Substitute into equation (2). Equation (2) contains y. Wherever y appears, write what it equals (inside a bracket). Now there is only one letter, x, so we can solve.
Equation (2): 3x − 2y = 7
3x − 2(7 − 2x) = 7
Expand the bracket and collect like terms. Multiply the bracket out, then add the x terms together and the plain numbers together.
Expand −2(7 − 2x):
(−2) × (−2x) = 4x
(−2) × 7 = −14
so −2(7 − 2x) = 4x − 14
The equation becomes: 4x − 14 + 3x = 7
Collect like terms: 7x − 14 = 7
Solve the equation for x. It is now a one-letter equation. Use inverse operations on both sides.
Add 14 to both sides: 7x = 21
Divide both sides by 7: x = 21 ÷ 7 = 3
Substitute x = 3 into the rearranged equation to find y. The rearranged equation y = 7 − 2x is the easiest to use.
y = 7 − 2x
y = 7 − 2 × 3
y = 1
Write the solution. Give both values as a pair.
x = 3, y = 1

Check. Put x = 3, y = 1 into equation (1): 2 × 3 + 1 = 6 + 1 = 7. It equals 7, as it should. ✓

Watch out. Put the expression in brackets when you substitute, and multiply the bracket by everything in front of it. A minus sign in front of a bracket changes every sign inside.

Watch out

Do not forget the brackets, and do not stop when you find the first letter. A solution has two values. Also, when you finish, put the values back into the other equation to check, not only into the one you rearranged.

14.7

Solving equations with graphs

Edexcel 1MA1: A18, A19
Grade 3-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Solving equations with graphsComing soon

In Chapter 12 you learned that the equation of a straight line, such as y = 2x + 1, is a rule that every point on the line obeys. Now suppose we draw two lines on the same grid. A point that is on both lines obeys both equations at once. So the point where the lines cross is the solution of the simultaneous equations. This is the graphical method.

The lines y = x + 1 and y = 7 − x cross at (3, 4). So the solution is x = 3, y = 4.

To draw a straight line make a table of values: choose three values of x, work out y for each, plot the points and join them with a ruler. If the equation is like x + y = 8, first rearrange it to y = 8 − x. Another quick way is to find where the line crosses the axes: put x = 0 to find where it crosses the y-axis, and y = 0 to find where it crosses the x-axis.

Graphs give approximate answers, because you can only read a graph to a certain accuracy. When the crossing point is at a grid corner, you can read it exactly, and you should always check by substituting into both equations. If two lines are parallel (same gradient) they never cross, so the simultaneous equations have no solution.

Solving quadratic equations with a graph. The solutions of x² − 4x + 1 = 0 are the values of x where the curve y = x² − 4x + 1 meets the line y = 0 (the x-axis). Read the x-values. These are usually not whole numbers, so give them to 1 decimal place. To solve x² − 2x − 3 = 2, draw the line y = 2 on the same graph and read the x-values where it meets the curve. In general, to solve an equation with a graph, plot each side as a graph and find where they cross.

The curve y = x² − 4x + 1 crosses the x-axis at about x = 0.3 and x = 3.7. These are approximate solutions of x² − 4x + 1 = 0.

Key ideas

The solution of simultaneous equations is the point where their graphs cross: (x, y).
Draw a line using a table of values (or the axis crossings). Rearrange to y = … first if you need to.
Parallel lines never cross, so there is no solution.
To solve f(x) = 0 with a graph, read the x-values where the curve meets the x-axis (y = 0).
To solve f(x) = k, draw the line y = k and read the x-values where it meets the curve.
Graph answers are approximate: give them to 1 decimal place when they are not whole numbers, and check by substituting.

Worked example 14.7.1

The graph shows the lines y = x + 1 and y = 7 − x. Use it to solve the simultaneous equations y = x + 1 and y = 7 − x.

Key words. The solution of two simultaneous equations is the pair (x, y) that makes both true. On a graph it is where the two lines intersect (cross).

Method. The solution of two simultaneous equations is the point where their graphs cross.

Find the point where the two lines cross. Every point on a line fits that line's equation. The one point that is on both lines fits both equations, so it is the solution of the simultaneous equations.
Look for the crossing point of the two lines.
Read the coordinates of the crossing point. The coordinates are written (x, y): read along the x-axis first, then up the y-axis.
The lines cross at (3, 4).
Write the solution. The x-coordinate is the value of x and the y-coordinate is the value of y.
x = 3, y = 4

Check. Put x = 3 into both equations. y = x + 1: y = 3 + 1 = 4. y = 7 − x: y = −3 + 7 = 4. Both give y = 4. ✓

Watch out. Give both coordinates and say which is x and which is y. Graph answers are only as accurate as your reading, so check by substituting.

Worked example 14.7.2

By drawing graphs, solve y = 2x − 1 and x + y = 8.

Method. Draw both straight lines on the same grid. The solution is the point where they cross.

Rearrange to the form y = mx + c (where necessary). It is much easier to find points when y is on its own.
x + y = 8 → subtract x from both sides → y = 8 − x
Make a table of values for line 1: y = 2x − 1. Choose some easy values of x, put each one into the equation and work out y. Two points are enough to draw a straight line, but a third point checks your arithmetic.
When x = 0: y = 2 × 0 − 1 = 0 − 1 = −1
When x = 2: y = 2 × 2 − 1 = 4 − 1 = 3
When x = 4: y = 2 × 4 − 1 = 8 − 1 = 7
When x = 6: y = 2 × 6 − 1 = 12 − 1 = 11
Make a table of values for line 2: x + y = 8. Choose some easy values of x, put each one into the equation and work out y. Two points are enough to draw a straight line, but a third point checks your arithmetic.
When x = 0: y = −0 + 8 = 0 + 8 = 8
When x = 2: y = −2 + 8 = 6
When x = 4: y = −4 + 8 = 4
When x = 6: y = −6 + 8 = 2
Plot the points and draw both lines. Plot each pair (x, y), then join the points of each line with a ruler. Extend the lines so that they cross.
Read off the crossing point. The point where the lines cross is on both lines, so it is the solution.
The lines cross at (3, 5).
x = 3, y = 5

Check. Check by substituting x = 3: line 1 gives y = 2 × 3 − 1 = 5 and line 2 gives y = −3 + 8 = 5. They agree. ✓

Watch out. Use a sharp pencil and a ruler, and extend the lines far enough to see where they cross.

Worked example 14.7.3

Explain why the simultaneous equations y = 2x + 1 and y = 2x + 4 have no solution.

Method. Compare the gradients. If two straight lines have the same gradient they are parallel.

Write the gradient of each line. In y = mx + c, the number m in front of x is the gradient.
y = 2x + 1 has gradient 2
y = 2x + 4 has gradient 2
Compare. The gradients are equal, so the lines slope the same way and stay the same distance apart. They never cross.
The lines are parallel.
Conclude. A solution would be a point on both lines, and there is no such point.
There is no solution.

Check. Algebra agrees: if 2x + 1 = 2x + 4 then 1 = 4, which is false. ✓

Watch out. Parallel lines have the same m but different c.

Worked example 14.7.4

The graph of y = x² − 4x + 1 is shown above. Use it to find estimates of the solutions of x² − 4x + 1 = 0.

Method. The solutions of x² − 4x + 1 = 0 are the x-values where the curve y = x² − 4x + 1 crosses the x-axis.

Find where the curve meets y = 0. The x-axis is the line y = 0, and an equation “expression = 0” asks where the expression is 0.
The curve crosses the x-axis twice: just right of 0, and just left of 4.
Read the x-values as accurately as you can. Each small square on the grid is 1 unit, so estimate the position between the grid lines. The first crossing is about a quarter of the way from 0 to 1. The second is about three-quarters of the way from 3 to 4.
x ≈ 0.3 and x ≈ 3.7
Write the answers to 1 decimal place. These are approximate answers because a graph cannot be read exactly.
x = 0.3 or x = 3.7 (to 1 d.p.)

Check. Substitute x = 0.3: 0.3² − 4 × 0.3 + 1 = 0.09 − 1.2 + 1 = −0.11, close to 0. Substitute x = 3.7: 3.7² − 4 × 3.7 + 1 = 13.69 − 14.8 + 1 = −0.11, close to 0. ✓

Watch out. Graph readings are only approximate, so any answer within about 0.1 of the true value is acceptable.

Worked example 14.7.5

Use the graph of y = x² − 2x − 3 to estimate the solutions of x² − 2x − 3 = 4.

Method. To solve a quadratic equation with a graph, draw a line that represents the other side, then read the x-values where the curve and the line meet.

Write the equation as a pair of graphs. The left side x² − 2x − 3 is the curve. The right side is 4. So we need the curve y = x² − 2x − 3 and the line y = 4.
Curve: y = x² − 2x − 3
Line: y = 4
Find where they meet.
The line y = 4 meets the curve at two points.
Read the x-values. Estimate the positions between grid lines.
x ≈ -1.8 and x ≈ 3.8

Check. Substitute x = 3.8: 3.8² − 2 × 3.8 − 3 = 14.44 − 7.6 − 3 = 3.84, close to 4. ✓

Watch out. Draw the correct horizontal line: for the equation = 4 the line is y = 4, not y = 0.

Watch out

Always read both coordinates and write the answer as x = …, y = … (not just a point). And remember a graph gives an approximate answer unless the crossing is exactly at a grid corner: check by substituting your values into both equations.

14.8

Forming simultaneous equations from words

Edexcel 1MA1: A19, A21
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Forming simultaneous equations from wordsComing soon

Simultaneous equations are used whenever there are two unknown quantities and two pieces of information about them: two shopping baskets, two journeys, two ages, two shapes. The skill is translating the words into algebra. Then you solve with elimination (section 14.5) or substitution (14.6).

Strategy.

Choose letters for the two unknowns and write down what each stands for, with units (for example “Let a be the cost of one adult ticket in pounds”).
Write equation (1) from the first piece of information, then equation (2) from the second. Check that both sides use the same units.
Solve the equations.
Answer the question in words, with units. The question may ask for something different from the letters, such as the total cost of a different basket.
Check your values in the original words, not just in your equations.

Useful translations: “the sum of” means add, “the difference between” means subtract, “a is 3 more than b” means a = b + 3 (or a − b = 3), “twice” means ×2, and the perimeter of a rectangle is 2l + 2w. For ages, remember that after n years each person is n years older. For a fixed charge plus a rate, the total is (fixed charge) + (rate × amount).

Key ideas

Two unknowns need two equations. Choose a letter for each unknown and say what it stands for, with units.
Turn each sentence of information into one equation. Keep the units the same.
Solve with elimination or substitution, then give the answer in words with units.
The question may not ask for the letters themselves. Read the last sentence again.
Check in the original sentences.

Worked example 14.8.1

2 adult tickets and 3 child tickets cost £31. 3 adult tickets and 2 child tickets cost £34. Work out the cost of one adult ticket and the cost of one child ticket.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
Let one adult ticket cost £a and one child ticket cost £c.
Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
2 adult tickets and 3 child tickets cost £31.
Total cost in pounds = (2 × a) + (3 × c).
2a + 3c = 31
Make equation (2) from the second sentence. The second piece of information gives a second equation.
3 adult tickets and 2 child tickets cost £34.
Total cost in pounds = (3 × a) + (2 × c).
3a + 2c = 34
Number the equations. We will keep referring to them, so call them (1) and (2).
(1) 2a + 3c = 31
(2) 3a + 2c = 34
Choose a letter to eliminate: a, and make the numbers match. The numbers in front of a are 2 and 3. The lowest common multiple of 2 and 3 is 6, so we will make both equal to 6. So multiply equation (1) by 3 and multiply equation (2) by 2. Multiply every term, including the number on the right-hand side.
Multiply to make the numbers in front of a match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
Equation (1) × 3: 3 × 2a, 3 × 3c on the left and 3 × 31 on the right.
(3) 6a + 9c = 93

Equation (2) × 2: 2 × 3a, 2 × 2c on the left and 2 × 34 on the right.
(4) 6a + 4c = 68
Subtract (4) from (3). The a terms are 6a and 6a. They have the same sign, so subtract the equations: 6a − 6a = 0, so a disappears. Do the same thing to the left sides and to the right sides.
(3) 6a + 9c = 93
(4) 6a + 4c = 68
a terms: 6a − 6a = 0 (they cancel)
c terms: 9c − 4c = 5c
Numbers: 93 − 68 = 25
5c = 25
Solve for c. 5c means 5 multiplied by c, so divide both sides by 5.
Divide both sides by 5: c = 25 ÷ 5 = 5
Substitute c = 5 into equation (1) to find a. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
Equation (1): 2a + 3c = 31
2a + 3 × 5 = 31
2a + 15 = 31
Subtract 15 from both sides: 2a = 16
Divide both sides by 2: a = 16 ÷ 2 = 8
Write the solution. Give both values. They belong together as a pair.
a = 8, c = 5
Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
Adult ticket: a = 8, so one adult ticket costs £8.
Child ticket: c = 5, so one child ticket costs £5.
One adult ticket costs £8 and one child ticket costs £5.

Check. Put a = 8, c = 5 into the other equation (equation (2), which we did not use to find a): 3 × 8 + 2 × 5 = 24 + 10 = 34. This equals 34, as it should. ✓

Watch out. Write what each letter stands for, including the units (pounds here). Check the answer in both sentences of the question.

Worked example 14.8.2

The sum of Mia’s age and her brother’s age is 31. Mia is 4 years older than twice her brother’s age. Find both ages.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
Let Mia be m years old and her brother be b years old.
Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
Their ages add up to 31.
Mia’s age is 4 more than twice her brother’s age: m = 2b + 4, which rearranges to m − 2b = 4.
m + b = 31
Make equation (2) from the second sentence. The second piece of information gives a second equation.
Mia is 4 years older than twice her brother’s age.
m = 2b + 4, so m − 2b = 4.
m − 2b = 4
Number the equations. We will keep referring to them, so call them (1) and (2).
(1) m + b = 31
(2) m − 2b = 4
Choose a letter to eliminate: m. To eliminate a letter, the numbers in front of it must be the same size in both equations. The numbers in front of m are 1 and 1: they are already the same size, so no multiplying is needed.
Subtract (2) from (1). The m terms are m and m. They have the same sign, so subtract the equations: m − m = 0, so m disappears. Do the same thing to the left sides and to the right sides.
(1) m + b = 31
(2) m − 2b = 4
m terms: m − m = 0 (they cancel)
b terms: b − (−2b) = 3b
Numbers: 31 − 4 = 27
3b = 27
Solve for b. 3b means 3 multiplied by b, so divide both sides by 3.
Divide both sides by 3: b = 27 ÷ 3 = 9
Substitute b = 9 into equation (1) to find m. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
Equation (1): m + b = 31
m + 9 = 31
m + 9 = 31
Subtract 9 from both sides: m = 22
Write the solution. Give both values. They belong together as a pair.
m = 22, b = 9
Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
Mia: m = 22 years.
Her brother: b = 9 years.
Mia is 22 and her brother is 9.

Check. Put m = 22, b = 9 into the other equation (equation (2), which we did not use to find m): 22 − 2 × 9 = 22 − 18 = 4. This equals 4, as it should. ✓

Watch out. Read the question again at the end: it may ask for something different from the letters you found (a total, a difference, or a different quantity).

Worked example 14.8.3

The perimeter of a rectangle is 38 cm. The length is 5 cm more than the width. Find the length and the width.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
Let the length be l cm and the width be w cm.
Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
The perimeter is two lengths and two widths: 2l + 2w = 38.
2l + 2w = 38
Make equation (2) from the second sentence. The second piece of information gives a second equation.
The length is 5 more than the width: l = w + 5, so l − w = 5.
l − w = 5
Number the equations. We will keep referring to them, so call them (1) and (2).
(1) 2l + 2w = 38
(2) l − w = 5
Choose a letter to eliminate: l, and make the numbers match. The numbers in front of l are 2 and 1. The lowest common multiple of 2 and 1 is 2, so we will make both equal to 2. So multiply equation (2) by 2. Multiply every term, including the number on the right-hand side.
Multiply to make the numbers in front of l match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
Equation (2) × 2: 2 × l, 2 × (−w) on the left and 2 × 5 on the right.
(3) 2l − 2w = 10
Subtract (3) from (1). The l terms are 2l and 2l. They have the same sign, so subtract the equations: 2l − 2l = 0, so l disappears. Do the same thing to the left sides and to the right sides.
(1) 2l + 2w = 38
(3) 2l − 2w = 10
l terms: 2l − 2l = 0 (they cancel)
w terms: 2w − (−2w) = 4w
Numbers: 38 − 10 = 28
4w = 28
Solve for w. 4w means 4 multiplied by w, so divide both sides by 4.
Divide both sides by 4: w = 28 ÷ 4 = 7
Substitute w = 7 into equation (2) to find l. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
Equation (2): l − w = 5
l − 7 = 5
l − 7 = 5
Add 7 to both sides: l = 12
Write the solution. Give both values. They belong together as a pair.
l = 12, w = 7
Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
Length = 12 cm, width = 7 cm.
The rectangle is 12 cm by 7 cm.

Check. Put l = 12, w = 7 into the other equation (equation (1), which we did not use to find l): 2 × 12 + 2 × 7 = 24 + 14 = 38. This equals 38, as it should. ✓

Watch out. Read the question again at the end: it may ask for something different from the letters you found (a total, a difference, or a different quantity).

Worked example 14.8.4

A boat travels 24 km downstream in 2 hours. It returns 24 km upstream in 3 hours. The boat’s own speed is b km/h and the speed of the current is c km/h. Find b and c.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
Downstream the current helps the boat, so its speed is b + c. Upstream the current slows the boat, so its speed is b − c.
Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
Downstream: speed = distance ÷ time = 24 ÷ 2 = 12 km/h.
b + c = 12
b + c = 12
Make equation (2) from the second sentence. The second piece of information gives a second equation.
Upstream: speed = 24 ÷ 3 = 8 km/h.
b − c = 8
b − c = 8
Number the equations. We will keep referring to them, so call them (1) and (2).
(1) b + c = 12
(2) b − c = 8
Choose a letter to eliminate: b. To eliminate a letter, the numbers in front of it must be the same size in both equations. The numbers in front of b are 1 and 1: they are already the same size, so no multiplying is needed.
Subtract (2) from (1). The b terms are b and b. They have the same sign, so subtract the equations: b − b = 0, so b disappears. Do the same thing to the left sides and to the right sides.
(1) b + c = 12
(2) b − c = 8
b terms: b − b = 0 (they cancel)
c terms: c − (−c) = 2c
Numbers: 12 − 8 = 4
2c = 4
Solve for c. 2c means 2 multiplied by c, so divide both sides by 2.
Divide both sides by 2: c = 4 ÷ 2 = 2
Substitute c = 2 into equation (1) to find b. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
Equation (1): b + c = 12
b + 2 = 12
b + 2 = 12
Subtract 2 from both sides: b = 10
Write the solution. Give both values. They belong together as a pair.
b = 10, c = 2
Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
Boat speed in still water: 10 km/h.
Current: 2 km/h.

Check. Put b = 10, c = 2 into the other equation (equation (2), which we did not use to find b): 10 − 2 = 10 − 2 = 8. This equals 8, as it should. ✓

Watch out. Read the question again at the end: it may ask for something different from the letters you found (a total, a difference, or a different quantity).

Watch out

Check that you have used both pieces of information, and that your final answer is what the question asked for. If the question asks for “the total cost of 1 adult and 1 child ticket”, finding the two separate prices is only part of the job. Also keep units consistent: do not mix pounds and pence in the same equation.

14.9

The quadratic formula

Edexcel 1MA1: A18
Grade 6-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: The quadratic formulaComing soon

Many quadratic equations cannot be factorised with whole numbers. For example x² + 3x − 5 = 0: you need two whole numbers that multiply to −5 and add to 3, and there are none. Yet the equation does have solutions (the graph crosses the axis). The quadratic formula finds them for any quadratic equation.

The formula. If ax² + bx + c = 0 then

x = −b ± √(b² − 4ac)2a

The ± sign means “plus or minus”: do the calculation once with + and once with −, giving two solutions. The number b² − 4ac is called the discriminant.

Where does it come from? It is just completing the square (section 14.10) done once for all quadratics. Start with ax² + bx + c = 0. Divide by a: x² + (b/a)x + c/a = 0. Complete the square: (x + b/2a)² − b²/4a² + c/a = 0. Move the numbers across: (x + b/2a)² = (b² − 4ac)/4a². Take the square root of both sides (remembering ±): x + b/2a = ±√(b² − 4ac)/2a. Subtract b/2a and you have the formula. So the formula is a short-cut that always works.

Using it. (1) Rearrange so that one side is 0 and the equation is in the form ax² + bx + c = 0. (2) Write down a, b and c with their signs. (3) Substitute into the formula, using brackets for negative numbers. (4) Work out b² − 4ac first. (5) Take the square root and finish. The answer may be a whole number, a fraction, a decimal rounded as the question asks, or an exact surd such as 3 ± √5 (see Chapter 6 for simplifying surds).

The discriminant tells you how many solutions. If b² − 4ac is positive there are two solutions. If it is 0 there is exactly one (a repeated root). If it is negative there are no real solutions because you cannot take the square root of a negative number: the graph never crosses the x-axis.

Key ideas

Quadratic formula: for ax² + bx + c = 0, x = (−b ± √(b² − 4ac)) ÷ 2a.
First rearrange to equal 0. Write down a, b, c with their signs and substitute in brackets.
The discriminant b² − 4ac: positive gives 2 solutions, zero gives 1, negative gives none.
Exact answers keep the surd, e.g. 3 ± √5. Decimal answers are rounded only at the very end.
The whole of the top, −b ± √…, is divided by 2a.

Worked example 14.9.1

Solve 2x² + 5x − 3 = 0 using the quadratic formula.

Key words. The discriminant is b² − 4ac. A surd is a square root that is not a whole number, like √5. An exact answer keeps the surd; a decimal answer is rounded.

Method. Rearrange to ax² + bx + c = 0, read off a, b, c, substitute them (in brackets) into the quadratic formula, work out the discriminant first, then give the answer in the form asked.

Write down the equation. The quadratic formula solves any equation in the form ax² + bx + c = 0.
2x² + 5x − 3 = 0
Read off a, b and c. a is the number in front of x², b is the number in front of x, and c is the number on its own. Each number keeps its sign (a minus sign belongs to the number after it). If there is no number in front of a letter, it is 1.
2x² + 5x − 3 = 0
a = 2, b = 5, c = −3
Write down the quadratic formula. This formula works for every quadratic equation. (It is proved by completing the square in general; see the start of this section.)
x = −b ± √(b² − 4ac)2a
Substitute a, b and c, using brackets. Brackets stop sign mistakes, especially when a number is negative, because (−5)² = 25 but −5² would be −25.
x = −5 ± √(5² − 4 × 2 × (−3))2 × 2
Work out the pieces one at a time. The part under the square root, b² − 4ac, is called the discriminant. Work it out separately and carefully; it decides what kind of answer we get.
b² = 5² = 25
4ac = 4 × 2 × (−3) = −24
b² − 4ac = 25 − (−24) = 49
−b = −5
2a = 2 × 2 = 4
Put the pieces into the formula. Now the formula only needs a square root and a division.
x = −5 ± √494
Find the square root of 49. 7 × 7 = 49, so √49 = 7. The ± sign means we do one calculation with + and one with −.
x = −5 ± 74
Work out the two answers. One calculation with the + sign and one with the − sign. Simplify each fraction.
With +: −5 + 74 = 24 = 12
With −: −5 − 74 = −124 = −3

Check. When x = −3: 2 × (−3)² + 5 × (−3) − 3 = 18 − 15 − 3 = 0. ✓
When x = 12: 2 × 12² + 5 × 12 − 3 = 12 + 52 − 3 = 0. ✓

Watch out. Put every number in brackets when you substitute. The whole top line is divided by 2a, not just the square root.

Worked example 14.9.2

Solve x² − 6x + 4 = 0. Give your answers in the form p ± √q.

Key words. The discriminant is b² − 4ac. A surd is a square root that is not a whole number, like √5. An exact answer keeps the surd; a decimal answer is rounded.

Method. Rearrange to ax² + bx + c = 0, read off a, b, c, substitute them (in brackets) into the quadratic formula, work out the discriminant first, then give the answer in the form asked.

Write down the equation. The quadratic formula solves any equation in the form ax² + bx + c = 0.
x² − 6x + 4 = 0
Read off a, b and c. a is the number in front of x², b is the number in front of x, and c is the number on its own. Each number keeps its sign (a minus sign belongs to the number after it). If there is no number in front of a letter, it is 1.
x² − 6x + 4 = 0
a = 1, b = −6, c = 4
Write down the quadratic formula. This formula works for every quadratic equation. (It is proved by completing the square in general; see the start of this section.)
x = −b ± √(b² − 4ac)2a
Substitute a, b and c, using brackets. Brackets stop sign mistakes, especially when a number is negative, because (−5)² = 25 but −5² would be −25.
x = −(−6) ± √((−6)² − 4 × 1 × 4)2 × 1
Work out the pieces one at a time. The part under the square root, b² − 4ac, is called the discriminant. Work it out separately and carefully; it decides what kind of answer we get.
b² = (−6)² = 36
4ac = 4 × 1 × 4 = 16
b² − 4ac = 36 − 16 = 20
−b = −(−6) = 6
2a = 2 × 1 = 2
Put the pieces into the formula. Now the formula only needs a square root and a division.
x = 6 ± √202
Simplify √20. 20 = 4 × 5 and 4 is a square number, so √20 = √4 × √5 = 2√5. (Look for the biggest square number that divides into 20.)
√20 = 2√5
Cancel the common factor in the fraction. Every number in the fraction (6, 2 and 2) can be divided by 2. Divide the top and the bottom by 2. You can only cancel a factor that divides every term on the top.
x = 6 ± 2√52 = 3 ± √5

Check. Substitute x = 0.76: 1 × (0.76)² + (−6) × (0.76) + 4 ≈ 0.018, which is very close to 0 (it is not exactly 0 only because 0.76 is rounded). ✓
Substitute x = 5.24: 1 × (5.24)² + (−6) × (5.24) + 4 ≈ 0.018, which is very close to 0 (it is not exactly 0 only because 5.24 is rounded). ✓

Watch out. Put every number in brackets when you substitute. The whole top line is divided by 2a, not just the square root.

Worked example 14.9.3

Solve 3x² + 4x − 2 = 0. Give your answers correct to 2 decimal places.

Key words. The discriminant is b² − 4ac. A surd is a square root that is not a whole number, like √5. An exact answer keeps the surd; a decimal answer is rounded.

Method. Rearrange to ax² + bx + c = 0, read off a, b, c, substitute them (in brackets) into the quadratic formula, work out the discriminant first, then give the answer in the form asked.

Write down the equation. The quadratic formula solves any equation in the form ax² + bx + c = 0.
3x² + 4x − 2 = 0
Read off a, b and c. a is the number in front of x², b is the number in front of x, and c is the number on its own. Each number keeps its sign (a minus sign belongs to the number after it). If there is no number in front of a letter, it is 1.
3x² + 4x − 2 = 0
a = 3, b = 4, c = −2
Write down the quadratic formula. This formula works for every quadratic equation. (It is proved by completing the square in general; see the start of this section.)
x = −b ± √(b² − 4ac)2a
Substitute a, b and c, using brackets. Brackets stop sign mistakes, especially when a number is negative, because (−5)² = 25 but −5² would be −25.
x = −4 ± √(4² − 4 × 3 × (−2))2 × 3
Work out the pieces one at a time. The part under the square root, b² − 4ac, is called the discriminant. Work it out separately and carefully; it decides what kind of answer we get.
b² = 4² = 16
4ac = 4 × 3 × (−2) = −24
b² − 4ac = 16 − (−24) = 40
−b = −4
2a = 2 × 3 = 6
Put the pieces into the formula. Now the formula only needs a square root and a division.
x = −4 ± √406
Simplify √40. 40 = 4 × 10 and 4 is a square number, so √40 = √4 × √10 = 2√10. (Look for the biggest square number that divides into 40.)
√40 = 2√10
Cancel the common factor in the fraction. Every number in the fraction (4, 2 and 6) can be divided by 2. Divide the top and the bottom by 2. You can only cancel a factor that divides every term on the top.
x = −4 ± 2√106 = −2 ± √103
Use a calculator and round to 2 decimal places. Type the whole top of the fraction in brackets, then divide by the bottom. Do the + and the − separately. Do not round until the end.
With +: x = −4 + 2√106 = 0.3874… ≈ 0.39
With −: x = −4 − 2√106 = −1.7208… ≈ −1.72

Check. Substitute x = −1.72: 3 × (−1.72)² + 4 × (−1.72) + (−2) ≈ −0.005, which is very close to 0 (it is not exactly 0 only because −1.72 is rounded). ✓
Substitute x = 0.39: 3 × (0.39)² + 4 × (0.39) + (−2) ≈ 0.016, which is very close to 0 (it is not exactly 0 only because 0.39 is rounded). ✓

Watch out. Put every number in brackets when you substitute. The whole top line is divided by 2a, not just the square root.

Worked example 14.9.4

Solve x² + 3x = 5. Give your answers correct to 2 decimal places.

Key words. The discriminant is b² − 4ac. A surd is a square root that is not a whole number, like √5. An exact answer keeps the surd; a decimal answer is rounded.

Method. Rearrange to ax² + bx + c = 0, read off a, b, c, substitute them (in brackets) into the quadratic formula, work out the discriminant first, then give the answer in the form asked.

Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
x² + 3x = 5
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 5 from both sides: x² + 3x − 5 = 0
Read off a, b and c. a is the number in front of x², b is the number in front of x, and c is the number on its own. Each number keeps its sign (a minus sign belongs to the number after it). If there is no number in front of a letter, it is 1.
x² + 3x − 5 = 0
a = 1, b = 3, c = −5
Write down the quadratic formula. This formula works for every quadratic equation. (It is proved by completing the square in general; see the start of this section.)
x = −b ± √(b² − 4ac)2a
Substitute a, b and c, using brackets. Brackets stop sign mistakes, especially when a number is negative, because (−5)² = 25 but −5² would be −25.
x = −3 ± √(3² − 4 × 1 × (−5))2 × 1
Work out the pieces one at a time. The part under the square root, b² − 4ac, is called the discriminant. Work it out separately and carefully; it decides what kind of answer we get.
b² = 3² = 9
4ac = 4 × 1 × (−5) = −20
b² − 4ac = 9 − (−20) = 29
−b = −3
2a = 2 × 1 = 2
Put the pieces into the formula. Now the formula only needs a square root and a division.
x = −3 ± √292
Write the exact answer. An answer with a square root left in it is called an exact answer (a surd). It is exactly right, with no rounding.
x = −3 ± √292
Use a calculator and round to 2 decimal places. Type the whole top of the fraction in brackets, then divide by the bottom. Do the + and the − separately. Do not round until the end.
With +: x = −3 + √292 = 1.1926… ≈ 1.19
With −: x = −3 − √292 = −4.1926… ≈ −4.19

Check. Substitute x = −4.19: 1 × (−4.19)² + 3 × (−4.19) + (−5) ≈ −0.014, which is very close to 0 (it is not exactly 0 only because −4.19 is rounded). ✓
Substitute x = 1.19: 1 × (1.19)² + 3 × (1.19) + (−5) ≈ −0.014, which is very close to 0 (it is not exactly 0 only because 1.19 is rounded). ✓

Watch out. Put every number in brackets when you substitute. The whole top line is divided by 2a, not just the square root.

Worked example 14.9.5

Show that x² + 2x + 5 = 0 has no real solutions.

Key words. The discriminant is b² − 4ac. A surd is a square root that is not a whole number, like √5. An exact answer keeps the surd; a decimal answer is rounded.

Method. Rearrange to ax² + bx + c = 0, read off a, b, c, substitute them (in brackets) into the quadratic formula, work out the discriminant first, then give the answer in the form asked.

Write down the equation. The quadratic formula solves any equation in the form ax² + bx + c = 0.
x² + 2x + 5 = 0
Read off a, b and c. a is the number in front of x², b is the number in front of x, and c is the number on its own. Each number keeps its sign (a minus sign belongs to the number after it). If there is no number in front of a letter, it is 1.
x² + 2x + 5 = 0
a = 1, b = 2, c = 5
Write down the quadratic formula. This formula works for every quadratic equation. (It is proved by completing the square in general; see the start of this section.)
x = −b ± √(b² − 4ac)2a
Substitute a, b and c, using brackets. Brackets stop sign mistakes, especially when a number is negative, because (−5)² = 25 but −5² would be −25.
x = −2 ± √(2² − 4 × 1 × 5)2 × 1
Work out the pieces one at a time. The part under the square root, b² − 4ac, is called the discriminant. Work it out separately and carefully; it decides what kind of answer we get.
b² = 2² = 4
4ac = 4 × 1 × 5 = 20
b² − 4ac = 4 − 20 = −16
−b = −2
2a = 2 × 1 = 2
Look at the discriminant. You cannot take the square root of a negative number and get an ordinary (real) number, because a number multiplied by itself is never negative.
−16 is negative, so there are no real solutions.

Watch out. Put every number in brackets when you substitute. The whole top line is divided by 2a, not just the square root.

Watch out

The most common slips are: (1) forgetting that −b is positive when b is negative, (2) writing −5² = −25 instead of (−5)² = 25, (3) dividing only the square root by 2a instead of the whole top line, and (4) forgetting the b² − 4ac sign change when c is negative: 4ac is negative so subtracting it adds.

14.10

Completing the square

Edexcel 1MA1: A18
Grade 6-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Completing the squareComing soon

Look at what happens when you expand a bracket that is squared: (x + 3)² = x² + 6x + 9. The expression x² + 6x + 9 is a perfect square. Completing the square means rewriting any quadratic x² + bx + c in the form (x + p)² + q, where the letter x appears only once. It is useful for three things: solving equations, finding the turning point of a graph, and proving things about quadratics.

To make x² + 6x into a square we split the 6x into two strips, each 3 by x. The big square then has side (x + 3), but it has a missing corner of area 3 × 3 = 9.

Why it works. In the picture, the area x² + 6x is an x by x square plus two strips of 3 by x. To fill the corner you need 9 more. So x² + 6x = (x + 3)² − 9. The rule is: halve the number in front of x, put it in the bracket, and subtract its square. So x² + bx + c = (x + b/2)² − (b/2)² + c. For x² + 6x + 2 this gives (x + 3)² − 9 + 2 = (x + 3)² − 7.

When the number in front of x² is not 1, take that number out of the first two terms first: 2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3.

Solving. To solve x² + 6x + 2 = 0, write (x + 3)² − 7 = 0, so (x + 3)² = 7, so x + 3 = ±√7, and x = −3 ± √7. Always include both the + and the − square roots.

Turning points. The curve y = (x + p)² + q has a squared bracket, which is never negative and is 0 when x = −p. So the curve’s lowest point is (−p, q). The x-coordinate has the opposite sign to the number in the bracket. For example y = x² − 6x + 5 = (x − 3)² − 4 has its minimum at (3, −4).

The curve y = x² − 6x + 5 = (x − 3)² − 4 has its minimum point (turning point) at (3, −4).

Key ideas

x² + bx + c = (x + b/2)² − (b/2)² + c: halve b, then subtract its square.
If the number in front of x² is a (not 1), factor a out of the x² and x terms first.
To solve: rearrange to (x + p)² = number, square-root both sides (±), then subtract p.
y = a(x + p)² + q has its turning point at (−p, q): a minimum if a > 0 and a maximum if a < 0.
Check by expanding the answer.

Worked example 14.10.1

Write x² + 6x + 2 in the form (x + p)² + q.

Key words. Completing the square means rewriting x² + bx + c as (x + p)² + q.

Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.

Write down the expression and what we want. We want to write it in the form a(x + p)² + q, with x appearing only once.
x² + 6x + 2
Halve the number in front of x. The bracket we want is (x + something)². When you expand (x + p)² you get x² + 2px + p², so the number in front of x is twice p. So p is half of it.
6 ÷ 2 = 3
The bracket is (x + 3)².
Expand the bracket to see what it really gives. The bracket gives the right x² and x terms, but it also gives an extra number that is not in our expression.
First: x × x = x²
Outer: x × 3 = 3x
Inner: 3 × x = 3x
Last: 3 × 3 = 9
(x + 3)² = x² + 6x + 9
Adjust the number on the end. Our expression has x² + 6x + 2, but the bracket gave an extra +9 at the end. So we must take that off again and then put on our own 2.
x² + 6x + 2 = (x + 3)² − 9 + 2
= (x + 3)² − 7

Check. Expand the answer: (x + 3)² − 7 gives x² + 6x + 2, the same as the original. ✓

Watch out. Halve the number in front of x, and then subtract its square. Do not add it.

Worked example 14.10.2

Write x² − 8x + 5 in the form (x + p)² + q.

Key words. Completing the square means rewriting x² + bx + c as (x + p)² + q.

Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.

Write down the expression and what we want. We want to write it in the form a(x + p)² + q, with x appearing only once.
x² − 8x + 5
Halve the number in front of x. The bracket we want is (x + something)². When you expand (x + p)² you get x² + 2px + p², so the number in front of x is twice p. So p is half of it.
−8 ÷ 2 = −4
The bracket is (x − 4)².
Expand the bracket to see what it really gives. The bracket gives the right x² and x terms, but it also gives an extra number that is not in our expression.
First: x × x = x²
Outer: x × (−4) = −4x
Inner: (−4) × x = −4x
Last: (−4) × (−4) = 16
(x − 4)² = x² − 8x + 16
Adjust the number on the end. Our expression has x² − 8x + 5, but the bracket gave an extra +16 at the end. So we must take that off again and then put on our own 5.
x² − 8x + 5 = (x − 4)² − 16 + 5
= (x − 4)² − 11

Check. Expand the answer: (x − 4)² − 11 gives x² − 8x + 5, the same as the original. ✓

Watch out. Halve the number in front of x, and then subtract its square. Do not add it.

Worked example 14.10.3

Write 2x² + 8x + 5 in the form a(x + p)² + q.

Key words. Completing the square means rewriting x² + bx + c as (x + p)² + q.

Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.

Write down the expression and what we want. We want to write it in the form a(x + p)² + q, with x appearing only once.
2x² + 8x + 5
Take the number 2 out of the x² and x terms only. We can only complete the square when the number in front of x² is 1. So factorise just the first two terms, and leave the number on the end alone.
2x² + 8x + 5 = 2(x² + 4x) + 5
Complete the square inside the bracket. Halve the number in front of x inside the bracket. Then subtract its square, because squaring the bracket adds it.
4 ÷ 2 = 2
x² + 4x = (x + 2)² − 4
Multiply the extra number by 2 and tidy up. The −4 is still inside the bracket that is multiplied by 2, so it must be multiplied by 2 to come outside. Then add the number on the end.
2[(x + 2)² − 4] + 5
= 2(x + 2)² − 8 + 5
= 2(x + 2)² − 3

Check. Expand the answer: 2(x + 2)² − 3 gives 2x² + 8x + 5, the same as the original. ✓

Watch out. Halve the number in front of x, and then subtract its square. Do not add it.

Worked example 14.10.4

Solve x² + 6x + 2 = 0 by completing the square. Give your answers in the form p ± √q.

Key words. Completing the square means rewriting x² + bx + c as (x + p)² + q. A surd is a square root that is not a whole number.

Method. Completing the square: write the quadratic as a(x + p)² + q, then rearrange to find x by square-rooting.

Write the equation. We will rewrite the left-hand side so that x appears only once, inside a bracket that is squared.
x² + 6x + 2 = 0
Halve the number in front of x. The bracket we want is (x + something)². When you expand (x + p)² you get x² + 2px + p², so the number in front of x is twice p. So p is half of it.
6 ÷ 2 = 3
The bracket is (x + 3)².
Expand the bracket to see what it really gives. The bracket gives the right x² and x terms, but it also gives an extra number that is not in our expression.
First: x × x = x²
Outer: x × 3 = 3x
Inner: 3 × x = 3x
Last: 3 × 3 = 9
(x + 3)² = x² + 6x + 9
Adjust the number on the end. Our expression has x² + 6x + 2, but the bracket gave an extra +9 at the end. So we must take that off again and then put on our own 2.
x² + 6x + 2 = (x + 3)² − 9 + 2
= (x + 3)² − 7
Write the equation using the new form. The expression is now equal to 0.
(x + 3)² − 7 = 0
Move the number to the right-hand side. Add 7 to both sides.
(x + 3)² = 7
Take the square root of both sides. There are two square roots of a positive number: a positive one and a negative one. For example 3 and −3 both square to make 9. So write ±.
x + 3 = ±√7
Subtract 3 from both sides. This leaves x on its own. This is the exact answer: it keeps the surd.
x = −3 ± √7

Check. Substitute x = −5.65: 1 × (−5.65)² + 6 × (−5.65) + 2 ≈ 0.023, which is very close to 0 (not exactly 0 only because −5.65 is rounded). ✓
Substitute x = −0.35: 1 × (−0.35)² + 6 × (−0.35) + 2 ≈ 0.023, which is very close to 0 (not exactly 0 only because −0.35 is rounded). ✓

Watch out. Square-rooting gives two answers (±). And the number inside the bracket is half of the number in front of x: remember to halve it, and to use the opposite sign when you solve.

Worked example 14.10.5

Solve 2x² + 12x + 6 = 0 by completing the square. Give your answers correct to 2 decimal places.

Key words. Completing the square means rewriting x² + bx + c as (x + p)² + q. A surd is a square root that is not a whole number.

Method. Completing the square: write the quadratic as a(x + p)² + q, then rearrange to find x by square-rooting.

Write the equation. We will rewrite the left-hand side so that x appears only once, inside a bracket that is squared.
2x² + 12x + 6 = 0
Take the number 2 out of the x² and x terms only. We can only complete the square when the number in front of x² is 1. So factorise just the first two terms, and leave the number on the end alone.
2x² + 12x + 6 = 2(x² + 6x) + 6
Complete the square inside the bracket. Halve the number in front of x inside the bracket. Then subtract its square, because squaring the bracket adds it.
6 ÷ 2 = 3
x² + 6x = (x + 3)² − 9
Multiply the extra number by 2 and tidy up. The −9 is still inside the bracket that is multiplied by 2, so it must be multiplied by 2 to come outside. Then add the number on the end.
2[(x + 3)² − 9] + 6
= 2(x + 3)² − 18 + 6
= 2(x + 3)² − 12
Write the equation using the new form. The expression is now equal to 0.
2(x + 3)² − 12 = 0
Move the number to the right-hand side. Add 12 to both sides.
2(x + 3)² = 12
Divide both sides by 2. This leaves the bracket squared on its own.
(x + 3)² = 12 ÷ 2 = 6
Take the square root of both sides. There are two square roots of a positive number: a positive one and a negative one. For example 3 and −3 both square to make 9. So write ±.
x + 3 = ±√6
Subtract 3 from both sides. This leaves x on its own. This is the exact answer: it keeps the surd.
x = −3 ± √6
Use a calculator to round to 2 decimal places. Work out the + and the − separately.
x = −3 + √6 = −0.5505… ≈ −0.55
x = −3 − √6 = −5.4495… ≈ −5.45

Check. Substitute x = −5.45: 2 × (−5.45)² + 12 × (−5.45) + 6 ≈ 0.005, which is very close to 0 (not exactly 0 only because −5.45 is rounded). ✓
Substitute x = −0.55: 2 × (−0.55)² + 12 × (−0.55) + 6 ≈ 0.005, which is very close to 0 (not exactly 0 only because −0.55 is rounded). ✓

Watch out. Square-rooting gives two answers (±). And the number inside the bracket is half of the number in front of x: remember to halve it, and to use the opposite sign when you solve.

Worked example 14.10.6

Write y = x² − 4x + 7 in completed-square form and find the coordinates of its turning point.

Key words. The turning point is the lowest point (minimum) or highest point (maximum) of a quadratic curve.

Method. Complete the square: y = a(x + p)² + q has its turning point at (−p, q).

Write down the expression. The turning point of the graph y = x² − 4x + 7 can be read straight from the completed-square form.
x² − 4x + 7
Halve the number in front of x. The bracket we want is (x + something)². When you expand (x + p)² you get x² + 2px + p², so the number in front of x is twice p. So p is half of it.
−4 ÷ 2 = −2
The bracket is (x − 2)².
Expand the bracket to see what it really gives. The bracket gives the right x² and x terms, but it also gives an extra number that is not in our expression.
First: x × x = x²
Outer: x × (−2) = −2x
Inner: (−2) × x = −2x
Last: (−2) × (−2) = 4
(x − 2)² = x² − 4x + 4
Adjust the number on the end. Our expression has x² − 4x + 7, but the bracket gave an extra +4 at the end. So we must take that off again and then put on our own 7.
x² − 4x + 7 = (x − 2)² − 4 + 7
= (x − 2)² + 3
Read off the turning point. A squared bracket is never negative, and it is exactly 0 when the bracket is 0. That happens when x − 2 = 0, i.e. x = 2. At that point the whole expression equals the number on the end, 3. Every other value of the bracket makes the expression bigger, so this is the minimum.
(x − 2)² + 3
Turning point = (2, 3), a minimum.

Check. Check using the graph: the curve should be symmetrical about x = 2. At x = 2: y = 2² − 4 × 2 + 7 = 4 − 8 + 7 = 3. ✓

Watch out. The x-coordinate of the turning point has the opposite sign to the number inside the bracket: (x + 3)² gives x = −3.

Watch out

When you complete the square you subtract the square of half the number (not add it), and when you read off the turning point from (x + 3)² − 7 the x-coordinate is −3, not 3. The number outside the bracket is q, with its own sign: (x + 3)² − 7 has y-coordinate −7.

14.11

Linear and quadratic simultaneous equations

Edexcel 1MA1: A19
Grade 7-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Linear and quadratic simultaneous equationsComing soon

So far both simultaneous equations have been straight lines. Now one equation can be a curve, such as y = x² − 4, or a circle, such as x² + y² = 25. Solving the pair means finding the points where the line meets the curve. A line can cross a parabola in two points, touch it at one point, or miss it completely.

The line y = x + 2 meets the curve y = x² − 4 at two points, (−2, 0) and (3, 5). Each point is one solution of the simultaneous equations.

The graph shows what the algebra must find, but graphs are only approximate. The algebraic method is substitution:

Rearrange the linear equation to make one letter the subject (for example y = x + 2).
Substitute it into the quadratic equation, in brackets. This gives an equation with only one letter. For y = x² − 4 and y = x + 2 we get x² − 4 = x + 2.
Rearrange to equal 0, factorise (or use the formula) and solve. This gives two values: x² − x − 6 = 0 so x = −2 or x = 3.
Put each value into the linear equation to find its y. This gives the pairs (−2, 0) and (3, 5).
Write the solutions as pairs.

Always use the linear equation to find the second value. If you use the quadratic you may get extra “answers” that are not solutions. For example in x² + y² = 25, x = 3 gives y = 4 or y = −4, but only one of them lies on the line.

Circles. The equation x² + y² = 25 is a circle with centre the origin and radius 5 (you will study circles in Chapter 25). The method is the same: substitute the linear equation into it. Be careful to square the whole bracket: (x + 1)² = x² + 2x + 1, not x² + 1.

How many solutions? After substitution you get a quadratic equation. If it has two solutions the line cuts the curve twice. If it has exactly one (a repeated root) the line is a tangent (it touches the curve). If the discriminant is negative there are no solutions and the line misses the curve.

The line y = x + 1 meets the circle x² + y² = 25 at (3, 4) and (−4, −3).

Key ideas

One equation is linear, one is quadratic (a parabola or a circle). The solutions are the points where they meet.
Make one letter the subject of the linear equation, substitute into the quadratic (in brackets), and solve.
There are usually two solutions. Find the second coordinate of each using the linear equation.
Write each solution as a pair (x, y), keeping the matching values together.
One repeated root means the line touches the curve; a negative discriminant means no solution.

Worked example 14.11.1

Solve the simultaneous equations y = x² − 4 and y = x + 2.

Key words. A linear equation gives a straight line. A quadratic equation involves a squared letter and gives a curve. Solving them simultaneously finds the points where the line meets the curve.

Method. Linear and quadratic simultaneous equations: rearrange the linear equation to make one letter the subject, substitute it into the quadratic equation, solve the resulting quadratic, then use the linear equation to find the other letter for each solution.

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) y = x² − 4
(2) y = x + 2
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = x + 2
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with x + 2:
x² − 4 = x + 2
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract x from both sides: x² − 4 − x = 2
Subtract 2 from both sides: x² − 4 − x − 2 = 0
Collect like terms: x² − x − 6 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -1 and c = -6.
We need two numbers that multiply to -6 (the last number) and add to -1 (the middle number). List the pairs that multiply to -6:
1 × -6 = -6, add: 1 + -6 = -5 ✗
-1 × 6 = -6, add: -1 + 6 = 5 ✗
2 × -3 = -6, add: 2 + -3 = -1 ✓ this pair works
Put the pair into two brackets: (x + 2)(x − 3).
Check by expanding: x² − 3x + 2x − 6 = x² − x − 6. ✓
x² − x − 6 = (x + 2)(x − 3)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x + 2)(x − 3) = 0
So (x − 3) = 0 or (x + 2) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 3 = 0
Add 3 to both sides: x = 3

x + 2 = 0
Subtract 2 from both sides: x = −2
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −2 or x = 3
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = x + 2 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −2: y = (−2) + 2 = 0
When x = 3: y = 3 + 2 = 5
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −2, y = 0 and x = 3, y = 5

Check. (−2, 0) in equation (1): x-part (−2)² − 4 = 4 − 4 = 0; this equals 0, the y value. ✓
(3, 5) in equation (1): x-part 3² − 4 = 9 − 4 = 5; this equals 5, the y value. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Worked example 14.11.2

Solve y = x² + x − 6 and 2x − y = 4.

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) y = x² + x − 6
(2) 2x − y = 4
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2): 2x − y = 4
Subtract 2x from both sides: −y = −2x + 4
Multiply both sides by −1 (change every sign): y = 2x − 4
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with 2x − 4:
x² + x − 6 = 2x − 4
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 2x from both sides: x² + x − 6 − 2x = −4
Add 4 to both sides: x² + x − 6 − 2x + 4 = 0
Collect like terms: x² − x − 2 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -1 and c = -2.
We need two numbers that multiply to -2 (the last number) and add to -1 (the middle number). List the pairs that multiply to -2:
1 × -2 = -2, add: 1 + -2 = -1 ✓ this pair works
Put the pair into two brackets: (x + 1)(x − 2).
Check by expanding: x² − 2x + 1x − 2 = x² − x − 2. ✓
x² − x − 2 = (x + 1)(x − 2)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x + 1)(x − 2) = 0
So (x − 2) = 0 or (x + 1) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 2 = 0
Add 2 to both sides: x = 2

x + 1 = 0
Subtract 1 from both sides: x = −1
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −1 or x = 2
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = 2x − 4 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −1: y = 2 × (−1) − 4 = −6
When x = 2: y = 2 × 2 − 4 = 0
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −1, y = −6 and x = 2, y = 0

Check. (−1, −6) in equation (1): x-part (−1)² + (−1) − 6 = 1 − 1 − 6 = −6; this equals −6, the y value. ✓
(2, 0) in equation (1): x-part 2² + 2 − 6 = 4 + 2 − 6 = 0; this equals 0, the y value. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Worked example 14.11.3

Solve x² + y² = 25 and y = x + 1.

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) x² + y² = 25
(2) y = x + 1
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = x + 1
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with x + 1. Put the expression in brackets:
x² + (x + 1)² = 25
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand (x + 1)(x + 1):
First: x × x = x²
Outer: x × 1 = x
Inner: 1 × x = x
Last: 1 × 1 = 1
so (x + 1)(x + 1) = x² + 2x + 1
The equation is now: 2x² + 2x + 1 = 25
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 25 from both sides: 2x² + 2x + 1 − 25 = 0
Collect like terms: 2x² + 2x − 24 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Look for a common factor. A common factor is something that divides into every term.
The terms are 2x², 2x, −24. The biggest number and the highest power of x that divide into all of them is 2.
Take it outside a bracket, and inside the bracket put each term divided by it:
2x² ÷ 2 = x²
2x ÷ 2 = x
−24 ÷ 2 = −12
So 2x² + 2x − 24 = 2(x² + x − 12).
Check by expanding: 2 × (x² + x − 12) = 2 × x², 2 × x, 2 × (−12) → 2x² + 2x − 24. ✓
Can the bracket be factorised again? Look at (x² + x − 12).
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 1 and c = -12.
We need two numbers that multiply to -12 (the last number) and add to 1 (the middle number). List the pairs that multiply to -12:
1 × -12 = -12, add: 1 + -12 = -11 ✗
-1 × 12 = -12, add: -1 + 12 = 11 ✗
2 × -6 = -12, add: 2 + -6 = -4 ✗
-2 × 6 = -12, add: -2 + 6 = 4 ✗
3 × -4 = -12, add: 3 + -4 = -1 ✗
-3 × 4 = -12, add: -3 + 4 = 1 ✓ this pair works
Put the pair into two brackets: (x − 3)(x + 4).
Check by expanding: x² + 4x − 3x − 12 = x² + x − 12. ✓
2x² + 2x − 24 = 2(x − 3)(x + 4)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
2(x − 3)(x + 4) = 0
So (x − 3) = 0 or (x + 4) = 0.
The number 2 in front can never be 0, so it gives no solution. Ignore it.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 3 = 0
Add 3 to both sides: x = 3

x + 4 = 0
Subtract 4 from both sides: x = −4
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −4 or x = 3
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = x + 1 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −4: y = (−4) + 1 = −3
When x = 3: y = 3 + 1 = 4
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −4, y = −3 and x = 3, y = 4

Check. (−4, −3) in equation (1): (−4)² + (−3)² = 16 + 9 = 25. ✓
(3, 4) in equation (1): 3² + 4² = 9 + 16 = 25. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Worked example 14.11.4

Solve x² + y² = 25 and x = 2y − 5. (This time x is already the subject.)

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) x² + y² = 25
(2) x = 2y − 5
Make x the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives x: x = 2y − 5
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace x in equation (1) with 2y − 5. Put the expression in brackets:
(2y − 5)² + y² = 25
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand (2y − 5)(2y − 5):
First: 2y × 2y = 4y²
Outer: 2y × (−5) = −10y
Inner: (−5) × 2y = −10y
Last: (−5) × (−5) = 25
so (2y − 5)(2y − 5) = 4y² − 20y + 25
The equation is now: 5y² − 20y + 25 = 25
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 25 from both sides: 5y² − 20y + 25 − 25 = 0
Collect like terms: 5y² − 20y = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Look for a common factor. A common factor is something that divides into every term.
The terms are 5y², −20y. The biggest number and the highest power of y that divide into all of them is 5y.
Take it outside a bracket, and inside the bracket put each term divided by it:
5y² ÷ 5y = y
−20y ÷ 5y = −4
So 5y² − 20y = 5y(y − 4).
Check by expanding: 5y × (y − 4) = 5y × y, 5y × (−4) → 5y² − 20y. ✓
5y² − 20y = 5y(y − 4)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
5y(y − 4) = 0
So (y − 4) = 0 or y = 0.
The number 5 in front can never be 0, so it gives no solution. Ignore it.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
y − 4 = 0
Add 4 to both sides: y = 4

y = 0
y = 0 is a solution.
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
y = 0 or y = 4
Find the other coordinate for each solution. We have the y values. Put each one into the linear equation x = 2y − 5 (never into the curve, which could give extra wrong pairs) to find the matching x value.
When y = 0: x = 2 × 0 − 5 = −5
When y = 4: x = 2 × 4 − 5 = 3
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −5, y = 0 and x = 3, y = 4

Check. (−5, 0) in equation (1): (−5)² + 0² = 25 + 0 = 25. ✓
(3, 4) in equation (1): 3² + 4² = 9 + 16 = 25. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Worked example 14.11.5

Show that the line y = 2x + 1 touches the curve y = x² + 2, and find the point where it touches.

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) y = x² + 2
(2) y = 2x + 1
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = 2x + 1
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with 2x + 1:
x² + 2 = 2x + 1
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 2x from both sides: x² + 2 − 2x = 1
Subtract 1 from both sides: x² + 2 − 2x − 1 = 0
Collect like terms: x² − 2x + 1 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -2 and c = 1.
We need two numbers that multiply to 1 (the last number) and add to -2 (the middle number). List the pairs that multiply to 1:
1 × 1 = 1, add: 1 + 1 = 2 ✗
-1 × -1 = 1, add: -1 + -1 = -2 ✓ this pair works
Put the pair into two brackets: (x − 1)(x − 1).
Check by expanding: x² − 1x − 1x + 1 = x² − 2x + 1. ✓
x² − 2x + 1 = (x − 1)²
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 1)² = 0
So (x − 1) = 0.
A bracket that appears twice gives the same answer twice, so there is only one solution from it (a repeated root).
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 1 = 0
Add 1 to both sides: x = 1
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = 1
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = 2x + 1 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = 1: y = 2 × 1 + 1 = 3
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = 1, y = 3

Check. (1, 3) in equation (1): x-part 1² + 2 = 1 + 2 = 3; this equals 3, the y value. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Watch out

The solutions are pairs. Do not write the two x values and then two unrelated y values. Also remember to find y from the linear equation, and to expand a squared bracket properly, for example (x − 3)² = x² − 6x + 9.

14.12

Quadratic equations with fractions

Edexcel 1MA1: A18, A4
Grade 7-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Quadratic equations with fractionsComing soon

Some equations contain the unknown in the denominator (the bottom of a fraction), for example x + 6/x = 5 or 2/x + 3/(x + 1) = 2. They look different, but multiplying through by the right expression turns them into ordinary quadratic equations. This uses the skills of algebraic fractions from Chapter 7 and of linear equations with fractions from Chapter 9.

Method.

Find the common denominator of all the fractions. If the bottoms are x and (x + 1), it is x(x + 1).
Multiply every term on both sides by it. Each fraction loses its bottom. Terms with no fraction, such as 5, are multiplied too: 5 becomes 5x(x + 1).
Expand any brackets and rearrange to equal 0.
Solve the quadratic by factorising or by the formula.
Check that no answer makes a denominator 0. You cannot divide by zero, so such an answer is rejected.

For example, for x + 6/x = 5, multiply every term by x: x² + 6 = 5x. Rearranged this is x² − 5x + 6 = 0, which gives (x − 2)(x − 3) = 0, so x = 2 or x = 3. Both are allowed because neither is 0.

Why does this work? Multiplying both sides of an equation by the same non-zero number keeps it balanced (Chapter 9). We are doing exactly that, with the number being the expression (common denominator). The only danger is a value of x that makes the expression 0, and that is why we check.

These equations also appear in problems about speed and time. If you travel 60 km at x km/h, the time taken is 60/x hours (time = distance ÷ speed). Two journeys give two fractions, and an equation for the total time is a quadratic after you clear the fractions.

Key ideas

Multiply every term by the common denominator to remove the fractions.
Expand brackets, rearrange to equal 0, then factorise or use the formula.
Reject any solution that makes an original denominator 0.
Time = distance ÷ speed leads to fractions with the speed in the denominator.
Check by substituting back into the original equation.

Worked example 14.12.1

Solve x + 6x = 5.

Key words. The denominator is the bottom of a fraction. The common denominator is a bottom that all the fractions can share. This works exactly like 1/2 + 1/3 = 3/6 + 2/6, except that the bottoms contain letters.

Method. Equations with letters in the denominators: (1) multiply every term by the common denominator to remove the fractions, (2) rearrange into a quadratic equal to 0, (3) solve it, (4) reject any answer that makes a denominator 0.

Write the equation. It has letters in the denominators (the bottoms of fractions). We will clear the fractions first.
x + 6x = 5
Find the common denominator of all the bottoms. The common denominator is the smallest expression that every bottom divides into. To remove the fractions we multiply every term, on both sides, by it.
x
Common denominator = x
Multiply every term on both sides by the common denominator. Doing the same to both sides keeps the equation balanced. Every term must be multiplied, including plain numbers.
x × x = x² (a term with no bottom is simply multiplied by x)
x × 6x = 6 (the bottom x divides into x exactly 1 time, so the bottom disappears and the top is multiplied by 1)
x × 5 = 5x (a term with no bottom is simply multiplied by x)
Write the new equation. There are no fractions left. It is a quadratic equation (it has a squared term).
x² + 6 = 5x
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 5x from both sides: x² + 6 − 5x = 0
Collect like terms: x² − 5x + 6 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -5 and c = 6.
We need two numbers that multiply to 6 (the last number) and add to -5 (the middle number). List the pairs that multiply to 6:
1 × 6 = 6, add: 1 + 6 = 7 ✗
-1 × -6 = 6, add: -1 + -6 = -7 ✗
2 × 3 = 6, add: 2 + 3 = 5 ✗
-2 × -3 = 6, add: -2 + -3 = -5 ✓ this pair works
Put the pair into two brackets: (x − 2)(x − 3).
Check by expanding: x² − 3x − 2x + 6 = x² − 5x + 6. ✓
x² − 5x + 6 = (x − 2)(x − 3)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 2)(x − 3) = 0
So (x − 3) = 0 or (x − 2) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 3 = 0
Add 3 to both sides: x = 3

x − 2 = 0
Add 2 to both sides: x = 2
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = 2 or x = 3
Check the answers are allowed. We multiplied by an expression containing the letter. An answer is only allowed if it does not make any of the original denominators equal to 0 (you cannot divide by 0).
Neither answer makes a denominator zero, so both are allowed.

Check. When x = 2: left side = 5 and right side = 5. ✓
When x = 3: left side = 5 and right side = 5. ✓

Watch out. Multiply every term, even the ones that are not fractions. After multiplying, rearrange to equal zero before factorising: never cancel an x from both sides.

Worked example 14.12.2

Solve 2x + 3x + 1 = 2.

Write the equation. It has letters in the denominators (the bottoms of fractions). We will clear the fractions first.
2x + 3x + 1 = 2
Find the common denominator of all the bottoms. The common denominator is the smallest expression that every bottom divides into. To remove the fractions we multiply every term, on both sides, by it.
x, x + 1
Common denominator = x(x + 1)
Multiply every term on both sides by the common denominator. Doing the same to both sides keeps the equation balanced. Every term must be multiplied, including plain numbers.
x(x + 1) × 2x = 2(x + 1) (the bottom x divides into x(x + 1) exactly (x + 1) times, so the bottom disappears and the top is multiplied by (x + 1))
x(x + 1) × 3x + 1 = 3x (the bottom x + 1 divides into x(x + 1) exactly x times, so the bottom disappears and the top is multiplied by x)
x(x + 1) × 2 = 2x(x + 1) (a term with no bottom is simply multiplied by x(x + 1))
Write the new equation. There are no fractions left. It is a quadratic equation (it has a squared term).
2(x + 1) + 3x = 2x(x + 1)
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand 2(x + 1):
2 × x = 2x
2 × 1 = 2
so 2(x + 1) = 2x + 2

Expand 2x(x + 1):
2x × x = 2x²
2x × 1 = 2x
so 2x(x + 1) = 2x² + 2x
The equation is now: 5x + 2 = 2x² + 2x
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 5x from both sides: 2 = 2x² + 2x − 5x
Subtract 2 from both sides: 0 = 2x² + 2x − 5x − 2
Collect like terms: 0 = 2x² − 3x − 2
Write it the other way round: 2x² − 3x − 2 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. A quadratic where the number in front of x² is 2, not 1. We use the split the middle term method.
Multiply the first and last numbers: 2 × (-2) = -4. We need two numbers that multiply to -4 and add to -3 (the middle number).
1 × -4 = -4, add: -3 ✓ this pair works
Split the middle term (-3x) into 1x and -4x: 2x² + 1x − 4x − 2.
Factorise each pair separately. First pair: 2x² + x = x(2x + 1). Second pair: −4x − 2 = −2(2x + 1).
The two brackets (2x + 1) are the same. That is how we know it has worked. Write: x(2x + 1) − 2(2x + 1).
Take the common bracket out: (2x + 1) multiplied by (x − 2).
Result: (x − 2)(2x + 1).
Check by expanding: (x − 2)(2x + 1) = 2x² − 3x − 2. ✓
2x² − 3x − 2 = (x − 2)(2x + 1)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 2)(2x + 1) = 0
So (x − 2) = 0 or (2x + 1) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 2 = 0
Add 2 to both sides: x = 2

2x + 1 = 0
Subtract 1 from both sides: 2x = −1
Divide both sides by 2: x = −1 ÷ 2 = −12
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −12 or x = 2
Check the answers are allowed. We multiplied by an expression containing the letter. An answer is only allowed if it does not make any of the original denominators equal to 0 (you cannot divide by 0).
Neither answer makes a denominator zero, so both are allowed.

Check. When x = −12: left side = 2 and right side = 2. ✓
When x = 2: left side = 2 and right side = 2. ✓

Watch out. Multiply every term, even the ones that are not fractions. After multiplying, rearrange to equal zero before factorising: never cancel an x from both sides.

Worked example 14.12.3

Solve 1x + 2x + 1 = 1. Give your answers in the form p ± √q, and also to 2 decimal places.

Write the equation. It has letters in the denominators (the bottoms of fractions). We will clear the fractions first.
1x + 2x + 1 = 1
Find the common denominator of all the bottoms. The common denominator is the smallest expression that every bottom divides into. To remove the fractions we multiply every term, on both sides, by it.
x, x + 1
Common denominator = x(x + 1)
Multiply every term on both sides by the common denominator. Doing the same to both sides keeps the equation balanced. Every term must be multiplied, including plain numbers.
x(x + 1) × 1x = (x + 1) (the bottom x divides into x(x + 1) exactly (x + 1) times, so the bottom disappears and the top is multiplied by (x + 1))
x(x + 1) × 2x + 1 = 2x (the bottom x + 1 divides into x(x + 1) exactly x times, so the bottom disappears and the top is multiplied by x)
x(x + 1) × 1 = x(x + 1) (a term with no bottom is simply multiplied by x(x + 1))
Write the new equation. There are no fractions left. It is a quadratic equation (it has a squared term).
(x + 1) + 2x = x(x + 1)
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand x(x + 1):
x × x = x²
x × 1 = x
so x(x + 1) = x² + x
The equation is now: 3x + 1 = x² + x
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 3x from both sides: 1 = x² + x − 3x
Subtract 1 from both sides: 0 = x² + x − 3x − 1
Collect like terms: 0 = x² − 2x − 1
Write it the other way round: x² − 2x − 1 = 0
Read off a, b and c. a is the number in front of x², b is the number in front of x, and c is the number on its own. Each number keeps its sign (a minus sign belongs to the number after it). If there is no number in front of a letter, it is 1.
x² − 2x − 1 = 0
a = 1, b = −2, c = −1
Write down the quadratic formula. This formula works for every quadratic equation. (It is proved by completing the square in general; see the start of this section.)
x = −b ± √(b² − 4ac)2a
Substitute a, b and c, using brackets. Brackets stop sign mistakes, especially when a number is negative, because (−5)² = 25 but −5² would be −25.
x = −(−2) ± √((−2)² − 4 × 1 × (−1))2 × 1
Work out the pieces one at a time. The part under the square root, b² − 4ac, is called the discriminant. Work it out separately and carefully; it decides what kind of answer we get.
b² = (−2)² = 4
4ac = 4 × 1 × (−1) = −4
b² − 4ac = 4 − (−4) = 8
−b = −(−2) = 2
2a = 2 × 1 = 2
Put the pieces into the formula. Now the formula only needs a square root and a division.
x = 2 ± √82
Simplify √8. 8 = 4 × 2 and 4 is a square number, so √8 = √4 × √2 = 2√2. (Look for the biggest square number that divides into 8.)
√8 = 2√2
Cancel the common factor in the fraction. Every number in the fraction (2, 2 and 2) can be divided by 2. Divide the top and the bottom by 2. You can only cancel a factor that divides every term on the top.
x = 2 ± 2√22 = 1 ± √2
Use a calculator and round to 2 decimal places. Type the whole top of the fraction in brackets, then divide by the bottom. Do the + and the − separately. Do not round until the end.
With +: x = 2 + 2√22 = 2.4142… ≈ 2.41
With −: x = 2 − 2√22 = −0.4142… ≈ −0.41
Check the answers are allowed. We multiplied by an expression containing the letter. An answer is only allowed if it does not make any of the original denominators equal to 0 (you cannot divide by 0).
Neither answer makes a denominator zero, so both are allowed.

Check. When x = −0.414: left side ≈ 1.000 and right side ≈ 1.000. ✓
When x = 2.414: left side ≈ 1.000 and right side ≈ 1.000. ✓

Watch out. Multiply every term, even the ones that are not fractions. After multiplying, rearrange to equal zero before factorising: never cancel an x from both sides.

Worked example 14.12.4

A cyclist rides 60 km at x km/h, then 60 km at (x + 10) km/h. The total time is 3.5 hours. Find x.

Form the equation. Time = distance ÷ speed. The first 60 km at x km/h takes 60/x hours. The second 60 km at (x + 10) km/h takes 60/(x + 10) hours. The total time is 3.5 hours, which is 7/2.
60/x + 60/(x + 10) = 7/2
Find the common denominator of all the bottoms. The common denominator is the smallest expression that every bottom divides into. To remove the fractions we multiply every term, on both sides, by it.
x, x + 10, 2
Common denominator = 2x(x + 10)
Multiply every term on both sides by the common denominator. Doing the same to both sides keeps the equation balanced. Every term must be multiplied, including plain numbers.
2x(x + 10) × 60x = 602(x + 10) (the bottom x divides into 2x(x + 10) exactly (2x + 20) times, so the bottom disappears and the top is multiplied by (2x + 20))
2x(x + 10) × 60x + 10 = 120x (the bottom x + 10 divides into 2x(x + 10) exactly 2x times, so the bottom disappears and the top is multiplied by 2x)
2x(x + 10) × 72 = 7x(x + 10) (the bottom 2 divides into 2x(x + 10) exactly (x² + 10x) times, so the bottom disappears and the top is multiplied by (x² + 10x))
Write the new equation. There are no fractions left. It is a quadratic equation (it has a squared term).
602(x + 10) + 120x = 7x(x + 10)
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand 60(2x + 20):
60 × 2x = 120x
60 × 20 = 1200
so 60(2x + 20) = 120x + 1200

Expand 7x(x + 10):
7x × x = 7x²
7x × 10 = 70x
so 7x(x + 10) = 7x² + 70x
The equation is now: 240x + 1200 = 7x² + 70x
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 240x from both sides: 1200 = 7x² + 70x − 240x
Subtract 1200 from both sides: 0 = 7x² + 70x − 240x − 1200
Collect like terms: 0 = 7x² − 170x − 1200
Write it the other way round: 7x² − 170x − 1200 = 0
Read off a, b and c. a is the number in front of x², b is the number in front of x, and c is the number on its own. Each number keeps its sign (a minus sign belongs to the number after it). If there is no number in front of a letter, it is 1.
7x² − 170x − 1200 = 0
a = 7, b = −170, c = −1200
Write down the quadratic formula. This formula works for every quadratic equation. (It is proved by completing the square in general; see the start of this section.)
x = −b ± √(b² − 4ac)2a
Substitute a, b and c, using brackets. Brackets stop sign mistakes, especially when a number is negative, because (−5)² = 25 but −5² would be −25.
x = −(−170) ± √((−170)² − 4 × 7 × (−1200))2 × 7
Work out the pieces one at a time. The part under the square root, b² − 4ac, is called the discriminant. Work it out separately and carefully; it decides what kind of answer we get.
b² = (−170)² = 28900
4ac = 4 × 7 × (−1200) = −33600
b² − 4ac = 28900 − (−33600) = 62500
−b = −(−170) = 170
2a = 2 × 7 = 14
Put the pieces into the formula. Now the formula only needs a square root and a division.
x = 170 ± √6250014
Find the square root of 62500. 250 × 250 = 62500, so √62500 = 250. The ± sign means we do one calculation with + and one with −.
x = 170 ± 25014
Work out the two answers. One calculation with the + sign and one with the − sign. Simplify each fraction.
With +: 170 + 25014 = 42014 = 30
With −: 170 − 25014 = −8014 = −407
Check the answers are allowed. We multiplied by an expression containing the letter. An answer is only allowed if it does not make any of the original denominators equal to 0 (you cannot divide by 0).
Neither answer makes a denominator zero, so both are allowed.
Decide which solution makes sense. A speed cannot be negative, so reject x = −40/7.
The first speed is 30 km/h (and the second is 30 + 10 = 40 km/h).

Check. When x = −407: left side = 72 and right side = 72. ✓
When x = 30: left side = 72 and right side = 72. ✓

Watch out. Multiply every term, even the ones that are not fractions. After multiplying, rearrange to equal zero before factorising: never cancel an x from both sides.

Watch out

Multiply every term, including the plain numbers, by the common denominator. A very common mistake is to multiply only the fractions. Another is to cancel an x from both sides of a quadratic, which loses a solution. And always check that your answers do not make a denominator 0.

Chapter summary

Skill	Method
Factorise x² + bx + c	Find two numbers that multiply to c and add to b: (x + p)(x + q). Take out a common factor first.
Difference of two squares	a² − b² = (a − b)(a + b)
Factorise ax² + bx + c (Higher)	Multiply a × c, find two numbers with that product that add to b, split the middle term and factorise in pairs.
Solve a quadratic by factorising	Rearrange to 0 on one side, factorise, set each bracket to 0 (null factor law), solve, check.
Problems leading to quadratics	Choose a letter, form an equation, solve, reject impossible answers, answer in words.
Simultaneous equations: elimination	Make one letter match in size, subtract if signs are the same and add if they are different, solve, substitute back, check.
Simultaneous equations: substitution	Make a letter the subject of one equation, substitute it (in brackets) into the other, solve, find the other letter, check.
Graphical solutions	The solution is where the graphs cross. Roots of a quadratic are where the curve meets y = 0.
Equations from words	Choose letters, write one equation per sentence, solve, answer in words with units.
Quadratic formula (Higher)	x = (−b ± √(b² − 4ac)) ÷ 2a; the discriminant decides how many solutions there are.
Completing the square (Higher)	x² + bx + c = (x + b/2)² − (b/2)² + c; turning point (−p, q).
Linear/quadratic simultaneous (Higher)	Substitute the linear equation into the quadratic, solve, find each second value from the linear equation, write pairs.
Quadratics with fractions (Higher)	Multiply every term by the common denominator, rearrange to 0, solve, reject any value that makes a denominator 0.

Can you…?

I can factorise quadratics of the form x² + bx + c and the difference of two squares.
I can solve quadratic equations by factorising, after rearranging so that one side is 0.
I can form a quadratic equation from an area or consecutive-number problem and reject impossible solutions.
I can solve simultaneous linear equations by elimination and by substitution.
I can solve simultaneous equations and estimate roots of quadratics using graphs.
I can form and solve simultaneous equations from a word problem.
Higher: I can factorise ax² + bx + c, use the quadratic formula, complete the square, solve linear/quadratic simultaneous equations and solve quadratic equations with fractions.

Key terms

Quadratic expression: An expression whose highest power is 2, such as x² + 5x + 6.
Quadratic equation: An equation of the form ax² + bx + c = 0.
Factorise: Write an expression as a product of brackets.
Null factor law: If a product is 0, at least one of its factors is 0.
Root: A solution of an equation. The roots of a quadratic are where its graph crosses the x-axis.
Difference of two squares: An expression a² − b², which factorises as (a − b)(a + b).
Simultaneous equations: Two or more equations that must be true at the same time.
Elimination: Adding or subtracting equations to remove one letter.
Substitution: Replacing a letter by an expression that equals it.
Discriminant: The expression b² − 4ac; its sign tells how many real solutions a quadratic has.
Completing the square: Writing x² + bx + c as (x + p)² + q.
Turning point: The lowest (minimum) or highest (maximum) point of a quadratic curve.
Surd: A square root that is not a whole number, such as √5; an exact answer keeps the surd.

Total 52 marks. Allow about 52 minutes.

← Chapter 13: Quadratic and other graphs Chapter 15: Ratio and proportion →

GCSE Maths Quadratic and Simultaneous Equations (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

You should already know

Where this leads

Exercise P Do you remember?

Factorising quadratic expressions

Key ideas

Watch out

Exercise 14A Factorising quadratic expressions

Solving quadratic equations by factorising

Key ideas

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Exercise 14B Solving quadratic equations by factorising

Rearranging first, and harder quadratics

Key ideas

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Exercise 14C Rearranging first, and harder quadratics

Problems leading to quadratic equations

Key ideas

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Exercise 14D Problems leading to quadratic equations

Simultaneous equations: elimination

Key ideas

Watch out

Exercise 14E Simultaneous equations: elimination

Simultaneous equations: substitution

Key ideas

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Exercise 14F Simultaneous equations: substitution

Solving equations with graphs

Key ideas

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Exercise 14G Solving equations with graphs

Forming simultaneous equations from words

Key ideas

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Exercise 14H Forming simultaneous equations from words

The quadratic formula

Key ideas

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Exercise 14I The quadratic formula

Completing the square

Key ideas

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Exercise 14J Completing the square

Linear and quadratic simultaneous equations

Key ideas

Watch out

Exercise 14K Linear and quadratic simultaneous equations

Quadratic equations with fractions

Key ideas

Watch out

Exercise 14L Quadratic equations with fractions

Chapter summary

Can you…?

Key terms

Exercise 14Z Check-up (non-calculator)

Exercise 14X Exam-style questions