Quadratic equations: factorising, formula, completing the square – GCSE Maths exam questions with mark scheme

1Grade 4–5Non-calculator3 marks

(a) Factorise $x^2 - 36$ [1]

(b) Solve $x^2 + 5x = 0$ [2]

(a) $(x + 6)(x - 6)$B1
(b) for $x(x + 5) = 0$M1
(b) $x = 0$ and $x = -5$A1
Answer: (a) $(x + 6)(x - 6)$ (b) $x = 0$, $x = -5$
Examiner tip: Never divide both sides by $x$ – you lose the solution $x = 0$.

2Grade 4–5Non-calculator3 marks

(a) Factorise $x^2 - 9x + 20$ [2]

(b) Hence solve $x^2 - 9x + 20 = 0$ [1]

(a) for $(x \pm 4)(x \pm 5)$, or brackets that expand to give two of the three terms correctlyM1
(a) $(x - 4)(x - 5)$A1
(b) $x = 4$ and $x = 5$ (ft their factorisation from (a))B1
Answer: (a) $(x - 4)(x - 5)$ (b) $x = 4$, $x = 5$
Examiner tip: Look for two numbers that multiply to $+20$ and add to $-9$ – both must be negative.

3Grade 4–5HigherNon-calculator3 marks

Solve $x^2 + 2x = 15$ [3]

for rearranging to $x^2 + 2x - 15 = 0$ (or completing the square to get $(x + 1)^2 = 16$); may be implied by a correct factorisation, correct formula substitution or $(x+1)^2 - 1 = 15$M1
for $(x + 5)(x - 3)$ (factors in either order) or correct substitution into the quadratic formula, e.g. $\frac{-2 \pm \sqrt{2^2 - 4(1)(-15)}}{2}$, or $x + 1 = \pm 4$M1
$x = -5$ and $x = 3$ (both needed; accept $x = 3$, $x = -5$ in either order)A1
Answer: $x = -5$, $x = 3$
Examiner tip: Rearrange to $x^2 + 2x - 15 = 0$ before factorising: writing $x(x + 2) = 15$ and then setting $x = 15$ or $x + 2 = 15$ is wrong, because each factor can only be set to zero when the other side is 0 (and guessing $3 \times 5$ misses $x = -5$).

4Grade 6–7HigherNon-calculator3 marks

Solve $3x^2 - 10x - 8 = 0$ by factorising. [3]

for $(3x \pm 2)(x \pm 4)$, or brackets that expand to give two of the three terms correctlyM1
$(3x + 2)(x - 4) = 0$A1
$x = -\tfrac23$ and $x = 4$A1
Answer: $x = -\tfrac{2}{3}$, $x = 4$
Examiner tip: When the coefficient of $x^2$ is not 1, expand your brackets to check the middle term before solving.

5Grade 6–7HigherCalculator3 marks

Solve $2x^2 + 5x - 4 = 0$

Give your solutions correct to 2 decimal places. [3]

for correct substitution into the quadratic formula, e.g. $\dfrac{-5 \pm \sqrt{5^2 - 4 \times 2 \times (-4)}}{2 \times 2}$ (allow one sign error in the substitution)M1
for $\dfrac{-5 \pm \sqrt{57}}{4}$M1
$x = 0.64$ and $x = -3.14$A1
Answer: $x = 0.64$, $x = -3.14$
Examiner tip: $-4ac = -4 \times 2 \times (-4) = +32$, so the discriminant is $25 + 32 = 57$ – a sign slip here is the most common error.

6Grade 6–7HigherNon-calculator4 marks

(a) Write $x^2 - 6x + 14$ in the form $(x - a)^2 + b$, where $a$ and $b$ are integers. [2]

(b) Hence write down the coordinates of the turning point of the curve $y = x^2 - 6x + 14$ [1]

(a) for $(x - 3)^2$ seen, or for $(x - 3)^2 - 9 + 14$M1
(a) $(x - 3)^2 + 5$A1
(b) $(3, 5)$ (ft their $a$ and $b$)B1
(c) e.g. $(x - 3)^2 \ge 0$ so $(x - 3)^2 + 5 \ge 5$, which is never $0$ (or: the minimum point is above the $x$-axis)C1
Answer: (a) $(x - 3)^2 + 5$ (b) $(3, 5)$ (c) $(x-3)^2 + 5 \ge 5 > 0$ for all $x$
Examiner tip: The turning point of $(x - a)^2 + b$ is $(a, b)$ – watch the sign: $(x - 3)^2$ gives $x = +3$.

7Grade 8–9HigherNon-calculator5 marks

(a) Write $2x^2 + 12x + 7$ in the form $a(x + b)^2 + c$, where $a$, $b$ and $c$ are integers. [3]

(b) Hence solve $2x^2 + 12x + 7 = 0$

Give your answers in the form $p \pm \dfrac{\sqrt{q}}{2}$, where $p$ and $q$ are integers. [2]

(a) for $2(x^2 + 6x) + 7$ or $2(x + 3)^2$ seenM1
(a) for $2(x + 3)^2 - 18 + 7$ oeM1
(a) $2(x + 3)^2 - 11$A1
(b) for $(x + 3)^2 = \dfrac{11}{2}$ (ft their (a))M1
(b) $x = -3 \pm \dfrac{\sqrt{22}}{2}$A1
Answer: (a) $2(x + 3)^2 - 11$ (b) $x = -3 \pm \dfrac{\sqrt{22}}{2}$
Examiner tip: When you take the 2 outside, the constant you subtract is $2 \times 3^2 = 18$, not $9$.

8Grade 8–9HigherCalculator6 marks

A rectangle has length $(2x + 3)$ cm and width $(x - 1)$ cm.

The area of the rectangle is $30\text{ cm}^2$.

(a) Show that $2x^2 + x - 33 = 0$ [2]

(b) Work out the perimeter of the rectangle. Give your answer correct to 3 significant figures. [4]

(a) for $(2x + 3)(x - 1) = 30$ expanded with at least 3 of the 4 terms correct: $2x^2 - 2x + 3x - 3$M1
(a) for $2x^2 + x - 3 = 30$ leading to $2x^2 + x - 33 = 0$ with no errorsA1
(b) for correct substitution into the quadratic formula: $\dfrac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-33)}}{2 \times 2}$M1
(b) $x = 3.819\ldots$ (negative root $-4.32\ldots$ rejected or ignored)A1
(b) for $2(2x + 3) + 2(x - 1)$ or $6x + 4$ with their positive $x$M1
(b) $26.9$ cm (accept $26.91$–$26.92$ before rounding)A1
Answer: (a) shown (b) $26.9$ cm
Examiner tip: Lengths must be positive – reject the negative root, and don't stop at $x$: the question asks for the perimeter.

9Grade 1–3Non-calculator4 marks

The diagram shows the graph of $y = x^2 - 2x - 3$ for values of $x$ from $-2$ to $4$

(a) Write down the coordinates of the point where the graph crosses the $y$-axis. [1]

(b) Use the graph to write down the solutions of $x^2 - 2x - 3 = 0$ [1]

(d) Use the graph to solve $x^2 - 2x - 3 = 5$ [1]

(a) $(0, -3)$ (accept $y = -3$ only if clearly identified as the intercept)B1
(b) $x = -1$ and $x = 3$ (both needed; do not accept coordinates $(-1, 0)$, $(3, 0)$ unless $x$-values are clearly given as the answer)B1
(c) $(1, -4)$B1
(d) $x = -2$ and $x = 4$ (both needed; e.g. from reading where $y = 5$ meets the curve)B1
Answer: (a) $(0, -3)$ (b) $x = -1$, $x = 3$ (c) $(1, -4)$ (d) $x = -2$, $x = 4$
Examiner tip: Solutions of $\ldots = 0$ are the $x$-values where the curve crosses the $x$-axis – give $x$-values, not the $y$-intercept.

10Grade 4–5FoundationNon-calculator3 marks

Hannah is solving the equation $(x - 5)(x + 2) = 0$

She writes:

$x = -5$ or $x = 2$

(a) Explain the mistake Hannah has made. [1]

(b) A quadratic equation has solutions $x = 3$ and $x = -7$

Write down this equation in the form $x^2 + bx + c = 0$, where $b$ and $c$ are integers. [2]

(a) e.g. she has used the wrong signs: $x - 5 = 0$ gives $x = 5$ and $x + 2 = 0$ gives $x = -2$ (an explanation that each bracket must equal zero, with the correct values $5$ and $-2$ stated)C1
(b) $(x - 3)(x + 7) = 0$ or $(x - 3)(x + 7)$ seenM1
(b) $x^2 + 4x - 21 = 0$ (accept $x^2 + 4x + (-21) = 0$; '$= 0$' must be present)A1
Answer: (a) The signs are wrong – the solutions are $x = 5$ and $x = -2$ (b) $x^2 + 4x - 21 = 0$
Examiner tip: Set each bracket equal to zero: the solution has the opposite sign to the number in the bracket.

11Grade 6–7HigherNon-calculator6 marks

The diagram shows a right-angled triangle.

The lengths of the sides, in cm, are $x$, $x + 7$ and $2x + 1$

(a) Show that $x^2 - 5x - 24 = 0$ [3]

(b) Work out the perimeter of the triangle. [3]

(a) $x^2 + (x + 7)^2 = (2x + 1)^2$ (Pythagoras with the hypotenuse $2x + 1$)M1
(a) correct expansion of both brackets: $x^2 + 14x + 49$ and $4x^2 + 4x + 1$M1
(a) correct rearrangement to $2x^2 - 10x - 48 = 0$ (or $0 = 2x^2 - 10x - 48$) and division by $2$ to reach $x^2 - 5x - 24 = 0$ with no errorsA1
(b) $(x - 8)(x + 3) = 0$ (or correct use of the formula)M1
(b) $x = 8$ (with $x = -3$ rejected or ignored)A1
(b) $40$ cm (sides $8$, $15$, $17$)A1
Answer: (a) shown (b) $40$ cm
Examiner tip: Square the whole bracket: $(2x + 1)^2 = 4x^2 + 4x + 1$, not $2x^2 + 1$ – and reject the negative length.

12Grade 6–7HigherCalculator4 marks

A ball is thrown upwards from a platform.

The height, $h$ metres, of the ball above the ground $t$ seconds after it is thrown is given by

$$h = 1.5 + 14t - 4.9t^2$$

(a) Work out the time it takes for the ball to hit the ground. Give your answer correct to 2 decimal places. [3]

(b) The equation $1.5 + 14t - 4.9t^2 = 0$ has a second solution. Explain why this solution is not used in part (a). [1]

(a) for $1.5 + 14t - 4.9t^2 = 0$ (or $4.9t^2 - 14t - 1.5 = 0$) with correct substitution into the quadratic formula, e.g. $\dfrac{-14 \pm \sqrt{14^2 - 4 \times (-4.9) \times 1.5}}{2 \times (-4.9)}$ (condone one sign error in the substitution if the formula is clearly correct)M1
(a) for $t = 2.96\ldots$ or $-0.103\ldots$ (either root correct, or $\sqrt{225.4}$ seen)A1
(a) $2.96$ seconds (accept $2.9605\ldots$ rounded correctly; do not accept two answers given as final)A1
(b) e.g. the other solution is negative ($t = -0.10$) and time cannot be negative / the ball had not been thrown thenC1
Answer: (a) $2.96$ s (b) The other solution, $t \approx -0.10$, is negative and time after the throw cannot be negative.
Examiner tip: Write the equation as $= 0$ before substituting – and keep the minus sign on $a = -4.9$ inside the formula.

13Grade 6–7HigherNon-calculator3 marks

Show that the solutions of $3x^2 + 2x - 2 = 0$ can be written as

$$x = \frac{-1 \pm \sqrt{7}}{3}$$ [3]

for correct substitution into the quadratic formula: $\dfrac{-2 \pm \sqrt{2^2 - 4 \times 3 \times (-2)}}{2 \times 3}$M1
for $\sqrt{28}$ simplified to $2\sqrt{7}$, e.g. $\dfrac{-2 \pm 2\sqrt{7}}{6}$ seenM1
for cancelling a common factor of $2$ to reach $\dfrac{-1 \pm \sqrt{7}}{3}$ with all steps shown and no errors (decimal working, e.g. $0.548\ldots$, scores M0)A1
Answer: Shown: $\dfrac{-2 \pm \sqrt{28}}{6} = \dfrac{-2 \pm 2\sqrt{7}}{6} = \dfrac{-1 \pm \sqrt{7}}{3}$
Examiner tip: In a 'show that', every step must be written: simplify the surd before cancelling – you can only cancel a factor that divides every term.

14Grade 8–9HigherNon-calculator6 marks

(a) Write $5 + 8x - 2x^2$ in the form $p - q(x - r)^2$, where $p$, $q$ and $r$ are integers. [3]

(b) Hence write down the coordinates of the turning point of the curve $y = 5 + 8x - 2x^2$, and state whether it is a maximum or a minimum. [1]

Give your answers in the form $a \pm \sqrt{b}$, where $a$ is an integer and $b$ is a fraction. [2]

(a) for taking out $-2$: $-2(x^2 - 4x) + 5$ or $-2(x^2 - 4x - \tfrac{5}{2})$M1
(a) for $-2[(x - 2)^2 - 4] + 5$ or $-2(x - 2)^2 + 8 + 5$ (completing the square inside the bracket)M1
(a) $13 - 2(x - 2)^2$ ($p = 13$, $q = 2$, $r = 2$)A1
(b) $(2, 13)$ and maximum (ft their $p$ and $r$; both parts needed)B1
(c) for $2(x - 2)^2 = 13$ or $(x - 2)^2 = \tfrac{13}{2}$ (ft their (a))M1
(c) $x = 2 \pm \sqrt{\tfrac{13}{2}}$ (accept $2 \pm \sqrt{6.5}$ or $2 \pm \tfrac{\sqrt{26}}{2}$; both signs needed)A1
Answer: (a) $13 - 2(x - 2)^2$ (b) $(2, 13)$, maximum (c) $x = 2 \pm \sqrt{\tfrac{13}{2}}$
Examiner tip: When the $x^2$ term is negative, factor out $-2$ first and multiply the $-4$ back out: $-2 \times -4 = +8$, not $-8$.

15Grade 8–9HigherCalculator5 marks

A rectangular lawn measures $12$ m by $8$ m.

A path of constant width $x$ metres is laid all the way around the lawn, as shown in the diagram.

The total area of the lawn and the path is $150\text{ m}^2$.

(a) Show that $2x^2 + 20x - 27 = 0$ [2]

(b) Work out the width of the path. Give your answer correct to 3 significant figures. [3]

(a) for $(12 + 2x)(8 + 2x) = 150$ (outer dimensions must be $12 + 2x$ and $8 + 2x$; or lawn area $96$ + path area $40x + 4x^2 = 150$)M1
(a) for $96 + 40x + 4x^2 = 150$ leading to $4x^2 + 40x - 54 = 0$ and division by $2$ to reach $2x^2 + 20x - 27 = 0$ with no errorsA1
(b) for correct substitution into the quadratic formula: $\dfrac{-20 \pm \sqrt{20^2 - 4 \times 2 \times (-27)}}{2 \times 2}$ (or a correct completing-the-square method reaching $(x + 5)^2 = 38.5$)M1
(b) for $x = 1.2048\ldots$ (negative root $-11.2\ldots$ may be seen)A1
(b) $1.20$ m (with the negative root rejected or not given as an answer; $1.2$ is not correct to 3 significant figures and scores A0 for this mark)A1
Answer: (a) shown (b) $1.20$ m
Examiner tip: The path goes on both sides, so each outer length is increased by $2x$, not $x$ – and $1.2$ is only 2 significant figures.

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