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GCSE Maths · EdexcelChapter 14 · Quadratics and simultaneous equations
GCSE Maths · Quadratic and Simultaneous Equations · Higher · Lesson 14.3

GCSE Maths: Rearranging first, and harder quadratics (Higher)

Part of the GCSE Maths quadratic and simultaneous equations chapter (Higher). Chapter overview · Foundation version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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14.3

Rearranging first, and harder quadratics

Edexcel 1MA1: A18
Grade 4-7

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

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.

  1. 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
  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 10 from both sides: x2 + 3x − 10 = 0
  3. 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)
  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 − 2)(x + 5) = 0
    So (x − 2) = 0 or (x + 5) = 0.
  5. 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
  6. 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.

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.

  1. 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
  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 2x from both sides: x2 − 2x = 15
    Subtract 15 from both sides: x2 − 2x − 15 = 0
  3. 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)
  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 − 5) = 0
    So (x − 5) = 0 or (x + 3) = 0.
  5. 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
  6. 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.

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.

  1. 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
  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 + 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
  3. 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
  4. 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)
  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 + 9) = 0
    So (x − 2) = 0 or (x + 9) = 0.
  6. 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
  7. 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.

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.

  1. 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
  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(x + 4):
    x × x = x2
    x × 4 = 4x
    so x(x + 4) = x2 + 4x
    The equation is now: x2 + 4x = 21
  3. 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
  4. 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)
  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 + 7) = 0
    So (x − 3) = 0 or (x + 7) = 0.
  6. 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
  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.

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.

  1. 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
  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 15 from both sides: 2x2 + x − 15 = 0
  3. 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)
  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)(2x − 5) = 0
    So (x + 3) = 0 or (2x − 5) = 0.
  5. 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
  6. 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.

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.

  1. 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
  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
  3. 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)
  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 + 2)(3x − 1) = 0
    So (x + 2) = 0 or (3x − 1) = 0.
  5. 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
  6. 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.

Exercise 14C  Rearranging first, and harder quadratics

Non-calculator
1
FSolve by rearranging and factorising.
(a)x2 + 2x = 15
(b)x2 + 5x = 14
(c)x2 − x = 12
(d)x2 = 3x + 10
(e)x2 = 7x − 12
(f)x2 + 6x = −8
2
FSolve. The letters appear on both sides.
(a)x2 + 3x = 2x + 6
(b)x2 − 5 = 4x
(c)x2 + 8 = 6x
(d)x2 − 2x = 3x − 6
(e)x2 + 4x = 2x + 3
(f)x2 − 3 = 2x
3
FExpand the brackets, then solve.
(a)x(x + 3) = 10
(b)x(x − 2) = 24
(c)(x + 1)(x + 2) = 6
(d)(x − 1)(x + 4) = 6
(e)(x + 3)(x + 4) = 30
(f)(x − 2)(x + 1) = 4
4
RAli solves (x + 1)(x + 3) = 15 like this: “x + 1 = 15, so x = 14; or x + 3 = 15, so x = 12.” Explain why this method is wrong and then solve the equation correctly.
5
PSShow that x(x + 5) = 14 can be written as x2 + 5x − 14 = 0, and then solve it.
6
RAmir solves x2 = 9x by dividing both sides by x, and gets “x = 9”. Find all the solutions of the equation and say what Amir missed.
7
PSA number squared, plus six times the number, makes 7. Find the number. (There are two possible answers.)
8
FSolve.
(a)2x2 + 5x = 3
(b)3x2 − 7x = −2
(c)6x2 + x = 2
(d)5x2 + 7x = −2
(e)4x2 − 9x = −2
(f)2x2 − x = 10
9
FSolve.
(a)2x2 + x = 15
(b)3x2 = 5x + 2
(c)6x2 = 7x + 3
(d)2x2 + 3x = 2x + 3
(e)(2x − 1)(x + 3) = 4
(f)x(2x + 5) = 3
10
RSam solves (x + 2)(x + 3) = (x + 2)(2x − 1) by dividing both sides by (x + 2). He finds only x = 4. Solve the equation properly and explain what Sam missed.

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