GCSE Maths: Problems leading to quadratic equations (Foundation)
Part of the GCSE Maths quadratic and simultaneous equations chapter (Foundation). Chapter overview · Higher version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Problems leading to quadratic equations
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).
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.
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. 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).
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. 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).
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. 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).
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.Exercise 14D Problems leading to quadratic equations
Non-calculatorKey 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. 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) = 72
- 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 = 72 - 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 72 from both sides: n2 + n − 72 = 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 = -72.
We need two numbers that multiply to -72 (the last number) and add to 1 (the middle number). List the pairs that multiply to -72:
1 × -72 = -72, add: 1 + -72 = -71 ✗
-1 × 72 = -72, add: -1 + 72 = 71 ✗
2 × -36 = -72, add: 2 + -36 = -34 ✗
-2 × 36 = -72, add: -2 + 36 = 34 ✗
3 × -24 = -72, add: 3 + -24 = -21 ✗
-3 × 24 = -72, add: -3 + 24 = 21 ✗
4 × -18 = -72, add: 4 + -18 = -14 ✗
-4 × 18 = -72, add: -4 + 18 = 14 ✗
6 × -12 = -72, add: 6 + -12 = -6 ✗
-6 × 12 = -72, add: -6 + 12 = 6 ✗
8 × -9 = -72, add: 8 + -9 = -1 ✗
-8 × 9 = -72, add: -8 + 9 = 1 ✓ this pair works
Put the pair into two brackets: (n − 8)(n + 9).
Check by expanding: n² + 9n − 8n − 72 = n2 + n − 72. ✓
n2 + n − 72 = (n − 8)(n + 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.(n − 8)(n + 9) = 0
So (n − 8) = 0 or (n + 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.n − 8 = 0
Add 8 to both sides: n = 8
n + 9 = 0
Subtract 9 from both sides: n = −9 - Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).n = −9 or n = 8
- Decide which solution makes sense. The numbers are positive, so reject n = −9.n = 8
- Find both numbers. The numbers are n and n + 1.8 and 8 + 1 = 9
The numbers are 8 and 9.
Check. Check: 8 × 9 = 72. ✓
Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).
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. 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) = 110
- 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 = 110 - 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 110 from both sides: n2 + n − 110 = 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 = -110.
We need two numbers that multiply to -110 (the last number) and add to 1 (the middle number). List the pairs that multiply to -110:
1 × -110 = -110, add: 1 + -110 = -109 ✗
-1 × 110 = -110, add: -1 + 110 = 109 ✗
2 × -55 = -110, add: 2 + -55 = -53 ✗
-2 × 55 = -110, add: -2 + 55 = 53 ✗
5 × -22 = -110, add: 5 + -22 = -17 ✗
-5 × 22 = -110, add: -5 + 22 = 17 ✗
10 × -11 = -110, add: 10 + -11 = -1 ✗
-10 × 11 = -110, add: -10 + 11 = 1 ✓ this pair works
Put the pair into two brackets: (n − 10)(n + 11).
Check by expanding: n² + 11n − 10n − 110 = n2 + n − 110. ✓
n2 + n − 110 = (n − 10)(n + 11) - 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 − 10)(n + 11) = 0
So (n − 10) = 0 or (n + 11) = 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 − 10 = 0
Add 10 to both sides: n = 10
n + 11 = 0
Subtract 11 from both sides: n = −11 - Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).n = −11 or n = 10
- Decide which solution makes sense. The numbers are positive, so reject n = −11.n = 10
- Find both numbers. The numbers are n and n + 1.10 and 10 + 1 = 11
The numbers are 10 and 11.
Check. Check: 10 × 11 = 110. ✓
Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).
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. 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) = 210
- 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 = 210 - 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 210 from both sides: n2 + n − 210 = 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 = -210.
We need two numbers that multiply to -210 (the last number) and add to 1 (the middle number). List the pairs that multiply to -210:
1 × -210 = -210, add: 1 + -210 = -209 ✗
-1 × 210 = -210, add: -1 + 210 = 209 ✗
2 × -105 = -210, add: 2 + -105 = -103 ✗
-2 × 105 = -210, add: -2 + 105 = 103 ✗
3 × -70 = -210, add: 3 + -70 = -67 ✗
-3 × 70 = -210, add: -3 + 70 = 67 ✗
5 × -42 = -210, add: 5 + -42 = -37 ✗
-5 × 42 = -210, add: -5 + 42 = 37 ✗
6 × -35 = -210, add: 6 + -35 = -29 ✗
-6 × 35 = -210, add: -6 + 35 = 29 ✗
7 × -30 = -210, add: 7 + -30 = -23 ✗
-7 × 30 = -210, add: -7 + 30 = 23 ✗
10 × -21 = -210, add: 10 + -21 = -11 ✗
-10 × 21 = -210, add: -10 + 21 = 11 ✗
14 × -15 = -210, add: 14 + -15 = -1 ✗
-14 × 15 = -210, add: -14 + 15 = 1 ✓ this pair works
Put the pair into two brackets: (n − 14)(n + 15).
Check by expanding: n² + 15n − 14n − 210 = n2 + n − 210. ✓
n2 + n − 210 = (n − 14)(n + 15) - 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 − 14)(n + 15) = 0
So (n − 14) = 0 or (n + 15) = 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 − 14 = 0
Add 14 to both sides: n = 14
n + 15 = 0
Subtract 15 from both sides: n = −15 - Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).n = −15 or n = 14
- Decide which solution makes sense. The numbers are positive, so reject n = −15.n = 14
- Find both numbers. The numbers are n and n + 1.14 and 14 + 1 = 15
The numbers are 14 and 15.
Check. Check: 14 × 15 = 210. ✓
Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).
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. 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 + 2) cm.
- Form an equation from the information. Area of a rectangle = length × width.x(x + 2) = 24
- 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 + 2):
x × x = x2
x × 2 = 2x
so x(x + 2) = x2 + 2x
The equation is now: x2 + 2x = 24 - 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 24 from both sides: x2 + 2x − 24 = 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 = -24.
We need two numbers that multiply to -24 (the last number) and add to 2 (the middle number). List the pairs that multiply to -24:
1 × -24 = -24, add: 1 + -24 = -23 ✗
-1 × 24 = -24, add: -1 + 24 = 23 ✗
2 × -12 = -24, add: 2 + -12 = -10 ✗
-2 × 12 = -24, add: -2 + 12 = 10 ✗
3 × -8 = -24, add: 3 + -8 = -5 ✗
-3 × 8 = -24, add: -3 + 8 = 5 ✗
4 × -6 = -24, add: 4 + -6 = -2 ✗
-4 × 6 = -24, add: -4 + 6 = 2 ✓ this pair works
Put the pair into two brackets: (x − 4)(x + 6).
Check by expanding: x² + 6x − 4x − 24 = x2 + 2x − 24. ✓
x2 + 2x − 24 = (x − 4)(x + 6) - 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 − 4)(x + 6) = 0
So (x − 4) = 0 or (x + 6) = 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
Add 4 to both sides: x = 4
x + 6 = 0
Subtract 6 from both sides: x = −6 - Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).x = −6 or x = 4
- Decide which solution makes sense. x is a width, so it cannot be negative. Reject x = −6.x = 4 cm
- Find the length and answer the question. The length is x + 2.Length = 4 + 2 = 6 cm
Width = 4 cm, length = 6 cm
Check. Check in the words: width 4 cm and length 6 cm differ by 2 cm and the area is 4 × 6 = 24 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).
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. 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 + 5) cm.
- Form an equation from the information. Area of a rectangle = length × width.x(x + 5) = 36
- 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 + 5):
x × x = x2
x × 5 = 5x
so x(x + 5) = x2 + 5x
The equation is now: x2 + 5x = 36 - 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 36 from both sides: x2 + 5x − 36 = 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 = -36.
We need two numbers that multiply to -36 (the last number) and add to 5 (the middle number). List the pairs that multiply to -36:
1 × -36 = -36, add: 1 + -36 = -35 ✗
-1 × 36 = -36, add: -1 + 36 = 35 ✗
2 × -18 = -36, add: 2 + -18 = -16 ✗
-2 × 18 = -36, add: -2 + 18 = 16 ✗
3 × -12 = -36, add: 3 + -12 = -9 ✗
-3 × 12 = -36, add: -3 + 12 = 9 ✗
4 × -9 = -36, add: 4 + -9 = -5 ✗
-4 × 9 = -36, add: -4 + 9 = 5 ✓ this pair works
Put the pair into two brackets: (x − 4)(x + 9).
Check by expanding: x² + 9x − 4x − 36 = x2 + 5x − 36. ✓
x2 + 5x − 36 = (x − 4)(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 − 4)(x + 9) = 0
So (x − 4) = 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 − 4 = 0
Add 4 to both sides: x = 4
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 = 4
- Decide which solution makes sense. x is a width, so it cannot be negative. Reject x = −9.x = 4 cm
- Find the length and answer the question. The length is x + 5.Length = 4 + 5 = 9 cm
Width = 4 cm, length = 9 cm
Check. Check in the words: width 4 cm and length 9 cm differ by 5 cm and the area is 4 × 9 = 36 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).
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. 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) = 40
- 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 = 40 - 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 40 from both sides: x2 + 3x − 40 = 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 = -40.
We need two numbers that multiply to -40 (the last number) and add to 3 (the middle number). List the pairs that multiply to -40:
1 × -40 = -40, add: 1 + -40 = -39 ✗
-1 × 40 = -40, add: -1 + 40 = 39 ✗
2 × -20 = -40, add: 2 + -20 = -18 ✗
-2 × 20 = -40, add: -2 + 20 = 18 ✗
4 × -10 = -40, add: 4 + -10 = -6 ✗
-4 × 10 = -40, add: -4 + 10 = 6 ✗
5 × -8 = -40, add: 5 + -8 = -3 ✗
-5 × 8 = -40, add: -5 + 8 = 3 ✓ this pair works
Put the pair into two brackets: (x − 5)(x + 8).
Check by expanding: x² + 8x − 5x − 40 = x2 + 3x − 40. ✓
x2 + 3x − 40 = (x − 5)(x + 8) - 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 − 5)(x + 8) = 0
So (x − 5) = 0 or (x + 8) = 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 + 8 = 0
Subtract 8 from both sides: x = −8 - Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).x = −8 or x = 5
- Decide which solution makes sense. x is a width, so it cannot be negative. Reject x = −8.x = 5 cm
- Find the length and answer the question. The length is x + 3.Length = 5 + 3 = 8 cm
Width = 5 cm, length = 8 cm
Check. Check in the words: width 5 cm and length 8 cm differ by 3 cm and the area is 5 × 8 = 40 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).
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. 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 + 1) cm.
- Form an equation from the information. Area of a rectangle = length × width.x(x + 1) = 56
- 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: x2 + x = 56 - 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 56 from both sides: x2 + x − 56 = 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 = -56.
We need two numbers that multiply to -56 (the last number) and add to 1 (the middle number). List the pairs that multiply to -56:
1 × -56 = -56, add: 1 + -56 = -55 ✗
-1 × 56 = -56, add: -1 + 56 = 55 ✗
2 × -28 = -56, add: 2 + -28 = -26 ✗
-2 × 28 = -56, add: -2 + 28 = 26 ✗
4 × -14 = -56, add: 4 + -14 = -10 ✗
-4 × 14 = -56, add: -4 + 14 = 10 ✗
7 × -8 = -56, add: 7 + -8 = -1 ✗
-7 × 8 = -56, add: -7 + 8 = 1 ✓ this pair works
Put the pair into two brackets: (x − 7)(x + 8).
Check by expanding: x² + 8x − 7x − 56 = x2 + x − 56. ✓
x2 + x − 56 = (x − 7)(x + 8) - 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 − 7)(x + 8) = 0
So (x − 7) = 0 or (x + 8) = 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 − 7 = 0
Add 7 to both sides: x = 7
x + 8 = 0
Subtract 8 from both sides: x = −8 - Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).x = −8 or x = 7
- Decide which solution makes sense. x is a width, so it cannot be negative. Reject x = −8.x = 7 cm
- Find the length and answer the question. The length is x + 1.Length = 7 + 1 = 8 cm
Width = 7 cm, length = 8 cm
Check. Check in the words: width 7 cm and length 8 cm differ by 1 cm and the area is 7 × 8 = 56 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).
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 the sentence as an equation. “The square of a number” is x2. “5 more than 4 times the number” is 4x + 5.x2 = 4x + 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 4x from both sides: x2 − 4x = 5
Subtract 5 from both sides: x2 − 4x − 5 = 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 = -5.
We need two numbers that multiply to -5 (the last number) and add to -4 (the middle number). List the pairs that multiply to -5:
1 × -5 = -5, add: 1 + -5 = -4 ✓ this pair works
Put the pair into two brackets: (x + 1)(x − 5).
Check by expanding: x² − 5x + 1x − 5 = x2 − 4x − 5. ✓
x2 − 4x − 5 = (x + 1)(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 + 1)(x − 5) = 0
So (x − 5) = 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 − 5 = 0
Add 5 to both sides: x = 5
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 = 5
Check. When x = −1: left side = (−1)2 = 1 = 1; right side = 4 × (−1) + 5 = 1. Both sides equal. ✓
When x = 5: left side = 52 = 25 = 25; right side = 4 × 5 + 5 = 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.
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 the sentence as an equation. A number is x. Its square is x2. Adding the number gives x2 + x.x2 + x = 42
- 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 42 from both sides: x2 + x − 42 = 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 = -42.
We need two numbers that multiply to -42 (the last number) and add to 1 (the middle number). List the pairs that multiply to -42:
1 × -42 = -42, add: 1 + -42 = -41 ✗
-1 × 42 = -42, add: -1 + 42 = 41 ✗
2 × -21 = -42, add: 2 + -21 = -19 ✗
-2 × 21 = -42, add: -2 + 21 = 19 ✗
3 × -14 = -42, add: 3 + -14 = -11 ✗
-3 × 14 = -42, add: -3 + 14 = 11 ✗
6 × -7 = -42, add: 6 + -7 = -1 ✗
-6 × 7 = -42, add: -6 + 7 = 1 ✓ this pair works
Put the pair into two brackets: (x − 6)(x + 7).
Check by expanding: x² + 7x − 6x − 42 = x2 + x − 42. ✓
x2 + x − 42 = (x − 6)(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 − 6)(x + 7) = 0
So (x − 6) = 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 − 6 = 0
Add 6 to both sides: x = 6
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 = 6
Check. When x = −7: left side = (−7)2 + (−7) = 49 − 7 = 42; right side = 42 = 42. Both sides equal. ✓
When x = 6: left side = 62 + 6 = 36 + 6 = 42; right side = 42 = 42. 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.
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. 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 w cm. Then the length is (w + 5) cm.
- Form an equation from the information. Area of a rectangle = length × width.w(w + 5) = 14
- 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 w(w + 5):
w × w = w2
w × 5 = 5w
so w(w + 5) = w2 + 5w
The equation is now: w2 + 5w = 14 - 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 14 from both sides: w2 + 5w − 14 = 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 w² + bw + c, with b = 5 and c = -14.
We need two numbers that multiply to -14 (the last number) and add to 5 (the middle number). List the pairs that multiply to -14:
1 × -14 = -14, add: 1 + -14 = -13 ✗
-1 × 14 = -14, add: -1 + 14 = 13 ✗
2 × -7 = -14, add: 2 + -7 = -5 ✗
-2 × 7 = -14, add: -2 + 7 = 5 ✓ this pair works
Put the pair into two brackets: (w − 2)(w + 7).
Check by expanding: w² + 7w − 2w − 14 = w2 + 5w − 14. ✓
w2 + 5w − 14 = (w − 2)(w + 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.(w − 2)(w + 7) = 0
So (w − 2) = 0 or (w + 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.w − 2 = 0
Add 2 to both sides: w = 2
w + 7 = 0
Subtract 7 from both sides: w = −7 - Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).w = −7 or w = 2
- Decide which solution makes sense. w is a width, so it cannot be negative. Reject w = −7.w = 2 cm
- Find the length and answer the question. The length is w + 5.Length = 2 + 5 = 7 cm
Width = 2 cm, length = 7 cm
Check. Check in the words: width 2 cm and length 7 cm differ by 5 cm and the area is 2 × 7 = 14 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).
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. 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.Perimeter = 34 m, so length + width = 17 m. Let the length be l m. Then the width is (17 − l) m.
- Form an equation from the information. Area = length × width = 60.l(17 − l) = 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 l(−l + 17):
l × (−l) = −l2
l × 17 = 17l
so l(−l + 17) = −l2 + 17l
The equation is now: −l2 + 17l = 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.Add l2 to both sides: 17l = 60 + l2
Subtract 17l from both sides: 0 = 60 + l2 − 17l
Collect like terms: 0 = l2 − 17l + 60
Write it the other way round: l2 − 17l + 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 l² + bl + c, with b = -17 and c = 60.
We need two numbers that multiply to 60 (the last number) and add to -17 (the middle number). List the pairs that multiply to 60:
1 × 60 = 60, add: 1 + 60 = 61 ✗
-1 × -60 = 60, add: -1 + -60 = -61 ✗
2 × 30 = 60, add: 2 + 30 = 32 ✗
-2 × -30 = 60, add: -2 + -30 = -32 ✗
3 × 20 = 60, add: 3 + 20 = 23 ✗
-3 × -20 = 60, add: -3 + -20 = -23 ✗
4 × 15 = 60, add: 4 + 15 = 19 ✗
-4 × -15 = 60, add: -4 + -15 = -19 ✗
5 × 12 = 60, add: 5 + 12 = 17 ✗
-5 × -12 = 60, add: -5 + -12 = -17 ✓ this pair works
Put the pair into two brackets: (l − 5)(l − 12).
Check by expanding: l² − 12l − 5l + 60 = l2 − 17l + 60. ✓
l2 − 17l + 60 = (l − 5)(l − 12) - 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.(l − 5)(l − 12) = 0
So (l − 12) = 0 or (l − 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.l − 12 = 0
Add 12 to both sides: l = 12
l − 5 = 0
Add 5 to both sides: l = 5 - Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).l = 5 or l = 12
- Decide which solutions make sense. Both answers are positive. They give the same rectangle, once as the length and once as the width.l = 5 gives width 12 m; l = 12 gives width 5 m
- Write the dimensions. The longer side is the length.The garden is 12 m by 5 m.
Check. Perimeter = 2 × (12 + 5) = 34 m and area = 12 × 5 = 60 m2. ✓
Watch out. When the equation is not in the form “x2 + …”, rearrange it first: 17l − l2 = 60 becomes l2 − 17l + 60 = 0.
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. 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 Sam's age be x years. His sister is 3 years younger, so she is (x − 3) years old.
- Form an equation from the information. The product of their ages is 130.x(x − 3) = 130
- 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 = 130 - 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 130 from both sides: x2 − 3x − 130 = 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 = -130.
We need two numbers that multiply to -130 (the last number) and add to -3 (the middle number). List the pairs that multiply to -130:
1 × -130 = -130, add: 1 + -130 = -129 ✗
-1 × 130 = -130, add: -1 + 130 = 129 ✗
2 × -65 = -130, add: 2 + -65 = -63 ✗
-2 × 65 = -130, add: -2 + 65 = 63 ✗
5 × -26 = -130, add: 5 + -26 = -21 ✗
-5 × 26 = -130, add: -5 + 26 = 21 ✗
10 × -13 = -130, add: 10 + -13 = -3 ✓ this pair works
Put the pair into two brackets: (x + 10)(x − 13).
Check by expanding: x² − 13x + 10x − 130 = x2 − 3x − 130. ✓
x2 − 3x − 130 = (x + 10)(x − 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.(x + 10)(x − 13) = 0
So (x − 13) = 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 − 13 = 0
Add 13 to both sides: x = 13
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 = 13
- Decide which solution makes sense. An age cannot be negative, so reject x = −10.x = 13
- Find both ages. Sam is x and his sister is x − 3.Sam: 13 years; sister: 13 − 3 = 10 years
Check. 13 × 10 = 130. ✓
Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).
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. 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 and the base (x + 4) cm.
- Form an equation from the information. Area = ½ × base × height = 30, so base × height = 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. Reject the negative height x = −10.x = 6
- Find the base. The base is x + 4.Base = 6 + 4 = 10 cm. Height 6 cm, base 10 cm.
Check. ½ × 10 × 6 = 30. ✓
Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).
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. 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.The border has width x cm on every side. So the whole frame is (12 + 2x) cm long and (8 + 2x) cm wide (the border is added at both ends).
- Form an equation from the information. The whole frame has area 140 cm2, which is length × width.(12 + 2x)(8 + 2x) = 140
- 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 (2x + 12)(2x + 8):
First: 2x × 2x = 4x2
Outer: 2x × 8 = 16x
Inner: 12 × 2x = 24x
Last: 12 × 8 = 96
so (2x + 12)(2x + 8) = 4x2 + 40x + 96
The equation is now: 4x2 + 40x + 96 = 140 - 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 140 from both sides: 4x2 + 40x + 96 − 140 = 0
Collect like terms: 4x2 + 40x − 44 = 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 4x2, 40x, −44. The biggest number and the highest power of x that divide into all of them is 4.
Take it outside a bracket, and inside the bracket put each term divided by it:
4x2 ÷ 4 = x2
40x ÷ 4 = 10x
−44 ÷ 4 = −11
So 4x2 + 40x − 44 = 4(x2 + 10x − 11).
Check by expanding: 4 × (x2 + 10x − 11) = 4 × x2, 4 × 10x, 4 × (−11) → 4x2 + 40x − 44. ✓
Can the bracket be factorised again? Look at (x2 + 10x − 11).
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 10 and c = -11.
We need two numbers that multiply to -11 (the last number) and add to 10 (the middle number). List the pairs that multiply to -11:
1 × -11 = -11, add: 1 + -11 = -10 ✗
-1 × 11 = -11, add: -1 + 11 = 10 ✓ this pair works
Put the pair into two brackets: (x − 1)(x + 11).
Check by expanding: x² + 11x − 1x − 11 = x2 + 10x − 11. ✓
4x2 + 40x − 44 = 4(x − 1)(x + 11) - 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 0.4(x − 1)(x + 11) = 0
So (x − 1) = 0 or (x + 11) = 0.
The number 4 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 − 1 = 0
Add 1 to both sides: x = 1
x + 11 = 0
Subtract 11 from both sides: x = −11 - Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).x = −11 or x = 1
- Decide which solution makes sense. The border width cannot be negative, so reject x = −11.x = 1
- Answer in words. The border width is x.The border is 1 cm wide.
Check. Whole frame: (12 + 2) by (8 + 2) = 14 by 10, and 14 × 10 = 140 cm2. ✓
Watch out. Every term has a common factor of 4: dividing by 4 makes the factorising easier, but the answers are the same.
Method. Try consecutive pairs of whole numbers and compare their products with 50. If 50 is skipped, no pair works.
- Write the products of consecutive numbers. The product of n and n + 1 gets bigger as n gets bigger, so we can look for where 50 would fit.5 × 6 = 30
6 × 7 = 42
7 × 8 = 56 - Compare with 50. 42 is less than 50 and 56 is more than 50, and there is no pair of consecutive whole numbers between 6, 7 and 7, 8.42 < 50 < 56
- Check with algebra. If n(n + 1) = 50 then n2 + n − 50 = 0. We would need two whole numbers that multiply to −50 and add to 1.Pairs for −50: (1, −50), (2, −25), (5, −10), (10, −5), (25, −2), (50, −1): their sums are −49, −23, −5, 5, 23, 49. None is 1.
- Conclude. No pair of consecutive whole numbers works.Two consecutive whole numbers cannot multiply to make 50.
Check. The pairs of consecutive numbers around 50 give 42 and 56, never 50. ✓
Watch out. A “show that” or “explain” question needs a clear conclusion in words.