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GCSE Maths · EdexcelChapter 14 · Quadratics and simultaneous equations
GCSE Maths · Quadratic and Simultaneous Equations · Higher · Lesson 14.12

GCSE Maths: Quadratic equations with fractions (Higher)

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

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14.12

Quadratic equations with fractions

Edexcel 1MA1: A18, A4
Grade 7-9

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.

  1. Find the common denominator of all the fractions. If the bottoms are x and (x + 1), it is x(x + 1).
  2. 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).
  3. Expand any brackets and rearrange to equal 0.
  4. Solve the quadratic by factorising or by the formula.
  5. 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.

  1. Write the equation. It has letters in the denominators (the bottoms of fractions). We will clear the fractions first.
    x + 6x = 5
  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
    Common denominator = x
  3. 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)
  4. Write the new equation. There are no fractions left. It is a quadratic equation (it has a squared term).
    x2 + 6 = 5x
  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 5x from both sides: x2 + 6 − 5x = 0
    Collect like terms: x2 − 5x + 6 = 0
  6. 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)
  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 − 2)(x − 3) = 0
    So (x − 3) = 0 or (x − 2) = 0.
  8. 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
  9. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
    x = 2 or x = 3
  10. 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.

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.

  1. Write the equation. It has letters in the denominators (the bottoms of fractions). We will clear the fractions first.
    2x + 3x + 1 = 2
  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)
  3. 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))
  4. 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)
  5. 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
  6. 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
  7. 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)
  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 − 2)(2x + 1) = 0
    So (x − 2) = 0 or (2x + 1) = 0.
  9. 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
  10. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
    x = −12 or x = 2
  11. 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.

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.

  1. Write the equation. It has letters in the denominators (the bottoms of fractions). We will clear the fractions first.
    1x + 2x + 1 = 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)
  3. 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))
  4. 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)
  5. 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
  6. 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
  7. 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
  8. 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
  9. 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
  10. 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
  11. Put the pieces into the formula. Now the formula only needs a square root and a division.
    x = 2 ± √82
  12. 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
  13. 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
  14. 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
  15. 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.

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.

  1. 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
  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)
  3. 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))
  4. 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)
  5. 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
  6. 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
  7. 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
  8. 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
  9. 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
  10. 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
  11. Put the pieces into the formula. Now the formula only needs a square root and a division.
    x = 170 ± √6250014
  12. 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
  13. 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
  14. 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.
  15. 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.

Exercise 14L  Quadratic equations with fractions

Non-calculator
1
FSolve.
(a)x + 12x = 7
(b)x − 6x = 1
(c)x + 8x = 6
(d)10x = x + 3
(e)x + 20x = 9
(f)18x = x − 3
2
FSolve 1x + 2x + 2 = 1.
3
FSolve 3x + 4x + 1 = 2.
4
FSolve 4x + 5x + 1 = 2.
5
FSolve x + 3x = 5. Give your answers correct to 2 decimal places.
6
FSolve 5x = x + 2. Give your answers correct to 2 decimal places.
7
FSolve 1x − 1 + 1x + 1 = 1. Give your answers correct to 2 decimal places.
8
PSA walker goes 24 km at x km/h and then returns the same 24 km at (x + 2) km/h. The total time for both journeys is 7 hours. Find x.
9
PSA number added to its reciprocal makes 103. Find the possible values of the number.
10
RShow that 1x + 2 + 1x + 3 = 712 can be written as 7x2 + 11x − 18 = 0. Then solve the equation.
11
RSolve x + 1x = 2 and explain why it has only one solution.

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