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A18

Solve quadratic equations (including those requiring rearrangement) algebraically by factorising, by completing the square and by using the quadratic formula; find approximate solutions using a graph

Quadratic equations

Worked answers, methods and verified real exam appearances for A18 on Edexcel GCSE Maths 1MA1.

Definition

Specification A18: solve quadratic equations, including those that require rearrangement, algebraically by factorising, by completing the square and by using the quadratic formula; find approximate solutions using a graph.

Explanation

  • Rearrange to ax2+bx+c=0ax^2+bx+c=0 first; expanding or clearing a fraction is often the first mark. Factorise when integer factors exist, then set each bracket to zero to get both roots.
  • Higher tier: completing the square writes x2−8x+3x^2-8x+3 as (x−4)2−13(x-4)^2-13, so (x−4)2=13(x-4)^2=13 and x=4±13x=4\pm\sqrt{13}; with a coefficient of x2x^2 above 11, take it out as a factor first.
  • The quadratic formula x=−b±b2−4ac2ax=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} works for any quadratic; write the substitution line with the signs shown, because sign errors are what the reports blame.
  • Give both solutions unless the context rules one out, in exact surd form when asked and otherwise rounded only at the end.
  • From a graph, the solutions of f(x)=0f(x)=0 are the xx-intercepts, and f(x)=kf(x)=k is read where the curve meets the line y=ky=k.

Worked example

1

Solve x2−8x+3=0x^2-8x+3=0. Give your answers in the form a±ba\pm\sqrt{b}, where aa and bb are integers.

(3)

(Total for Question 1 is 3 marks)

Working

  1. 1.Complete the square: x2−8x+3=(x−4)2−16+3=(x−4)2−13x^2-8x+3=(x-4)^2-16+3=(x-4)^2-13.
  2. 2.Set it equal to zero and isolate the square: (x−4)2=13(x-4)^2=13.
  3. 3.Take both square roots: x−4=±13x-4=\pm\sqrt{13}.
  4. 4.Add 44 to each: x=4±13x=4\pm\sqrt{13}.

Answer: x=4±13x=4\pm\sqrt{13}

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkAdditional guidance
1for (x−4)2−13=0(x-4)^2-13=0 or (x−4)2−16+3(x-4)^2-16+3, or for correct substitution into the quadratic formula, eg 8±64−122\dfrac{8\pm\sqrt{64-12}}{2}M1
for x−4=±13x-4=\pm\sqrt{13} or 8±522\dfrac{8\pm\sqrt{52}}{2}M1oe
for 4±134\pm\sqrt{13}A1cao; both solutions required

Common mistakes

  • Don't take only the positive square root and lose one solution.
  • Don't substitute bb into the formula without its sign, so −b-b is evaluated incorrectly.
  • Don't round a surd before the final answer and lose accuracy.

Mark-scheme wording to use

  • “M1 For (x − 3)^2 – 9 – 8 (= 0) or (x =) (−(−6) ± √((−6)^2 − 4(1)(−8)))/(2(1)) allow sign error for b M1 For x – 3 = ± √17 or x = (6 ± √68)/2 A1 cao” — Pearson 1MA1, Specimen 2 Paper 1H, Q17
  • “M1 Substitute into quadratic formula - allow sign errors M1 Evaluate as far as (5 ± √13)/2” — Pearson 1MA1, Sample Paper 3H, Q11
  • “M1 for factorisation eg (x ± 3)(x ± 5) or (x + a)(x + b) where ab = – 15 or a + b = – 2 or for substitution into quadratic formula … M1 for (x + 3)(x – 5) or for (2 ± √64)/2 A1 cao” — Pearson 1MA1, November 2024 Paper 1F, Q26

What the examiner wants

Signs under control. The June 2024 Paper 2H report says completing the square with a coefficient of x2x^2 greater than 11 is always a struggle, with credit for taking out the factor of 22 but very few completing it, and that where the quadratic formula was used it was invariably sign errors that led to incorrect solutions.

Exam tip

Write the substitution into the quadratic formula before evaluating; this is normally where the first method mark is earned.

Worked practice

Q1
Tier 1 · Easy

1

Solve x2+2x−35=0x^2+2x-35=0 by factorising.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkMark scheme
1
  • x=5x=5 or x=−7x=-7
2x2+2x−35=(x+7)(x−5)x^2+2x-35=(x+7)(x-5). By the zero-product rule, x+7=0x+7=0 or x−5=0x-5=0, so x=−7x=-7 or x=5x=5.
Q2
Tier 2 · Standard

2

Higher only: Solve x2+14x+5=0x^2+14x+5=0 by completing the square. Give exact answers.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
QuestionAnswerMarkMark scheme
2
  • x=−7±211x=-7\pm2\sqrt{11}
4x2+14x+5=(x+7)2−44x^2+14x+5=(x+7)^2-44. Hence (x+7)2=44(x+7)^2=44, so x+7=±44=±211x+7=\pm\sqrt{44}=\pm2\sqrt{11} and x=−7±211x=-7\pm2\sqrt{11}.
Q3
Tier 3 · Hard

3

Higher only: The curves y=x2y=x^2 and y=5x+1y=5x+1 intersect twice. Use the quadratic formula to find the exact xx-coordinates, then give the values a graph should show to two decimal places.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
QuestionAnswerMarkMark scheme
3
  • x=5±292x=\frac{5\pm\sqrt{29}}{2}
  • x≈−0.19x\approx-0.19 or x≈5.19x\approx5.19
5At an intersection, x2=5x+1x^2=5x+1, so x2−5x−1=0x^2-5x-1=0. The formula gives x=5±(−5)2−4(1)(−1)2=5±292x=\frac{5\pm\sqrt{(-5)^2-4(1)(-1)}}{2}=\frac{5\pm\sqrt{29}}{2}. These are approximately −0.19-0.19 and 5.195.19, matching the graph's intersection coordinates.
Q4
Tier 1 · Easy

4

Solve x2=64x^2=64.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
QuestionAnswerMarkMark scheme
4
  • x=−8x=-8 or x=8x=8
2Both 828^2 and (−8)2(-8)^2 equal 6464. Therefore x=−8x=-8 or x=8x=8.
Q5
Tier 2 · Standard

5

Solve x2−7x=44x^2-7x=44 by factorising.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
QuestionAnswerMarkMark scheme
5
  • x=−4x=-4 or x=11x=11
3Rearrange to get x2−7x−44=0x^2-7x-44=0. Factorise: (x−11)(x+4)=0(x-11)(x+4)=0. Therefore x=−4x=-4 or x=11x=11.
Q6
Tier 3 · Hard

6

Solve (x−4)(x+7)=12(x-4)(x+7)=12 by factorising.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
QuestionAnswerMarkMark scheme
6
  • x=−8x=-8 or x=5x=5
3Expand and rearrange: x2+3x−28=12x^2+3x-28=12, so x2+3x−40=0x^2+3x-40=0. Factorising gives (x+8)(x−5)=0(x+8)(x-5)=0, hence x=−8x=-8 or x=5x=5.
Q7
Tier 2 · Standard

7

Solve (x−3)2=49(x-3)^2=49.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
QuestionAnswerMarkMark scheme
7
  • x=−4x=-4 or x=10x=10
3Taking both square roots gives x−3=7x-3=7 or x−3=−7x-3=-7. Therefore x=10x=10 or x=−4x=-4.
Q8
Tier 3 · Hard

8

Higher only: Use the quadratic formula to solve 2x2+5x−13=02x^2+5x-13=0. Give your answers correct to 22 decimal places.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
QuestionAnswerMarkMark scheme
8
  • x=1.59x=1.59 or x=−4.09x=-4.09
4With a=2a=2, b=5b=5 and c=−13c=-13, the formula gives x=−5±52−4(2)(−13)4=−5±1294x=\dfrac{-5\pm\sqrt{5^2-4(2)(-13)}}{4}=\dfrac{-5\pm\sqrt{129}}{4}. The values are 1.589454…1.589454\ldots and −4.089454…-4.089454\ldots, which round to 1.591.59 and −4.09-4.09. Each root is 0.004454…0.004454\ldots, or 0.4450.445 final-digit units, from the nearest 22-decimal-place rounding boundary.
Q9
Tier 3 · Hard

9

Higher only: Solve (x+4)(x−1)=2x(x−3)(x+4)(x-1)=2x(x-3). Give both solutions correct to 22 decimal places.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
QuestionAnswerMarkMark scheme
9
  • x=8.53x=8.53 or x=0.47x=0.47
4Expanding both sides gives x2+3x−4=2x2−6xx^2+3x-4=2x^2-6x. Rearranging gives x2−9x+4=0x^2-9x+4=0. The quadratic formula gives x=9±652x=\dfrac{9\pm\sqrt{65}}{2}, so x=8.5311…x=8.5311\ldots or x=0.4688…x=0.4688\ldots, which round to 8.538.53 and 0.470.47.
Q10
Tier 3 · Hard

10

Higher only: Solve 2x(x−3)=5−x2x(x-3)=5-x using the quadratic formula. Give both solutions in exact form.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
QuestionAnswerMarkMark scheme
10
  • x=5±654x=\dfrac{5\pm\sqrt{65}}{4}
4Expand and rearrange to obtain 2x2−5x−5=02x^2-5x-5=0. With a=2a=2, b=−5b=-5 and c=−5c=-5, the quadratic formula gives x=5±(−5)2−4(2)(−5)4=5±654x=\dfrac{5\pm\sqrt{(-5)^2-4(2)(-5)}}{4}=\dfrac{5\pm\sqrt{65}}{4}.

Verified exam appearances

SeriesPaperQuestionMarksCalculatorTierLinks
2019-111HQ186Non-calculatorHigherQPMS
2022-111HQ195Non-calculatorHigherQPMS
2021-112FQ246AllowedFoundationQPMS
2019-063HQ205AllowedHigherQPMS
2024-062HQ56AllowedHigherQPMS
2024-111FQ263Non-calculatorFoundationQPMS
2023-061HQ245Non-calculatorHigherQPMS
2019-113HQ174AllowedHigherQPMS
2022-061HQ194Non-calculatorHigherQPMS
2024-112HQ215AllowedHigherQPMS
2019-111HQ215Non-calculatorHigherQPMS
2021-111HQ83Non-calculatorHigherQPMS
2022-063HQ225AllowedHigherQPMS
2019-111HQ144Non-calculatorHigherQPMS
2021-112FQ273AllowedFoundationQPMS
2021-113HQ164AllowedHigherQPMS
2021-111HQ224Non-calculatorHigherQPMS
2024-061HQ215Non-calculatorHigherQPMS
2022-112HQ193AllowedHigherQPMS
2023-113HQ177AllowedHigherQPMS
2024-111HQ135Non-calculatorHigherQPMS
2022-062HQ225AllowedHigherQPMS
2024-113HQ204AllowedHigherQPMS
2024-063HQ207AllowedHigherQPMS
2019-061HQ174Non-calculatorHigherQPMS
2023-063HQ214AllowedHigherQPMS

Other points in A Algebra · sequences and functions

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