Solving Quadratics

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More questions on this topic: GCSE Algebra: Solving Quadratics (154) IGCSE Algebra: Solving Quadratics (106) AS Pure: Quadratics (179) A2 Pure: Quadratics (105)

AQA

AQA · Old spec

June 2025 Paper 1 Q23

AQACurrent spec5 marksInequalities Solving Quadratics

23 The diagram shows a trapezium.

All lengths are in centimetres.

Trapezium with parallel sides 3x − 8 and x + 5 and perpendicular height 2x; not drawn accurately

The area of the trapezium is less than 22 cm\(^2\)

Work out the greatest possible range of values of \(x\).

Give your answer in the form \(\quad m \lt x \lt n \quad\) where \(m\) and \(n\) are both positive numbers. [5 marks]

Mark scheme

Answer	Mark	Comments
(Area \(=\)) \(\dfrac{1}{2} \times (x + 5 + 3x - 8) \times 2x\) or \(4x^2 - 3x\)	M1
\(4x^2 - 3x - 22 \lt 0\)	M1dep	oe needs to be a quadratic \(\lt 0\) condone \(4x^2 - 3x \lt 22\) if completing the square
\((4x - 11)(x + 2)\) or \(\dfrac{3 \pm \sqrt{9 + 352}}{8}\) or \(4\left[\left(x - \dfrac{3}{8}\right)^2 - \dfrac{361}{64}\right]\)	M1	must be from the correct quadratic oe could imply first 2 M marks if \(\lt 0\) seen
\((k \lt)\ x \lt 2\dfrac{3}{4}\)	A1	oe for \(2\dfrac{3}{4}\) where \(k \lt 2\dfrac{3}{4}\) or \(k\) could be blank could imply M3A1
\(2\dfrac{2}{3} \lt x\ \ (\lt k)\)	B1	oe for \(2\dfrac{2}{3}\) where \(k \gt 2\dfrac{2}{3}\) or \(k\) could be blank

Additional guidance

Inequalities changed to \(=\) without being recovered correctly will have a maximum mark of M2	M1M0M1A0B0
Inequalities changed to \(=\) then recovered can score full marks
Condone \((4x - 11)(4x + 8)\ (\lt 0)\) for third M mark

June 2025 Paper 1 Q11

AQACurrent spec4 marksFactorising Solving Quadratics

11 Solve \(\quad x^3 + 10x^2 + 8x = 0\)

Where appropriate give your answers in surd form.

Simplify your answers where possible. [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1: quadratic formula
\(x(x^2 + 10x + 8)\ \ (= 0)\) or \(x = 0\)	M1
\((x =)\ \dfrac{-10 \pm \sqrt{\left(10^2 - 4(\times 1) \times 8\right)}}{2(\times 1)}\)	M1	oe correct substitution into formula from correct quadratic
\((x =)\ \dfrac{-10 \pm \sqrt{100 - 32}}{2}\) or \(\dfrac{-10 \pm \sqrt{68}}{2}\)	A1	oe simplification eg \(-5 \pm \dfrac{\sqrt{68}}{2}\) dep on second M mark only implied by final A1
\((x =)\ 0,\ -5 \pm \sqrt{17}\)	A1	could be written as two separate surds.
Alternative method 2: completes the square
\(x(x^2 + 10x + 8)\ \ (= 0)\) or \(x = 0\)	M1
\(x\left[(x + 5)^2 \ldots\ldots\right]\)	M1	oe eg may not include the common factor of \(x\) at this stage must be from correct quadratic
\(x\left[(x + 5)^2 - 17\right]\)	A1	oe eg \((x + 5)^2 = 17\) dep on second M mark only condone common factor of \(x\) not being there at this stage
\((x =)\ 0,\ -5 \pm \sqrt{17}\)	A1	could be written as two separate surds.

Additional guidance

If \(x = 0\) stated as final answer then it doesn’t need to appear in the working
If \(x = 0\) not stated on answer line then second A mark cannot be awarded: eg \(x = 0\) is seen early in the working with everything else correct \(x = 0\) or \(x(x^2 + 10x + 8)\) not seen in the working but with everything else correct	M2A1A0 M0M1A1A0
Factor theorem used correctly to get complete answer	M2A2
Factor theorem used without complete answer	M0A0

June 2024 Paper 1 Q21

AQACurrent spec5 marksIndices Solving Quadratics

\[\left(2^{x + 3}\right)^{4x + 1} = 8^{x - 1}\]

Work out the possible values of \(x\). [5 marks]

Mark scheme

Answer	Mark	Comments
\(2^{(4x + 1)(x + 3)}\) or \(2^{4x^2 + 13x + 3}\)	M1	lhs
\(\left(2^3\right)^{x - 1}\) or \(2^{3(x - 1)}\) or \(2^{3x - 3}\)	M1	rhs
\(4x^2 + 12x + x + 3 = 3x - 3\) or \(4x^2 + 10x + 6\ (= 0)\) or \(2x^2 + 5x + 3\ (= 0)\)	M1dep	oe with expanded brackets dependent on both M marks \(4x^2 + 12x + x + 3\) implies first M mark \(3x - 3\) implies second M mark
\((2x + 3)(x + 1)\ (= 0)\) or if quadratic formula used \(\dfrac{-5 \pm \sqrt{5^2 - 24}}{4}\) or better	M1dep	oe factorised
\(-1\) and \(-1\dfrac{1}{2}\) or \(\dfrac{-3}{2}\) or \(-1.5\)	A1

Additional guidance

Both answers correct without working will get all marks	M4A1
Only one solution correct without working scores nothing	M0A0
No oe for A mark
Possible to make both sides powers of 8. Mark to the equivalent of this MS

June 2024 Paper 1 Q20

AQACurrent spec4 marksDrawing & Using Graphs Solving Quadratics

20 Here is the graph of \(\quad y = -x^2 + 3x + 6 \quad\) for \(\quad -2 \leqslant x \leqslant 5\)

Graph of y = −x² + 3x + 6 for −2 ⩽ x ⩽ 5 on a grid

By drawing a suitable straight line, work out approximate solutions to

\[x^2 - \frac{7}{2}x - 3 = 0\]

[4 marks]

Mark scheme

Answer	Mark	Comments
\((y =)\ \dfrac{-x}{2} + 3\)	B2	oe eg \(2y = -x + 6\) can be implied from correct line B1 for \((y =)\ \dfrac{x}{2} - 3\) can be implied from drawing this line or \(m = -\dfrac{1}{2}\) or \(c = +3\) not from incorrect working for either \(m\) or \(c\)
\(y = \dfrac{-x}{2} + 3\) accurately drawn on diagram	M1	ft their stated linear function in \(x\) accuracy is half a square at \(x = 0\) and \(x = 4\) needs to be a straight line (mark intention)
\([-0.6, -0.8]\) and [4.1, 4.3]	A1	dependent on B2M1

Additional guidance

Factorising by use of the quadratic formula alone	B0M0A0
Ignore any y values for A1
Do not allow M mark for \(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\) linear function drawn
Graph of \(x^2 - \dfrac{7}{2}x - 3 = 0\)	B0M0A0

June 2024 Paper 1 Q14

AQACurrent spec4 marksCompleting the Square Solving Quadratics

(a) Write \(\quad 2x^2 - 16x + 7 \quad\) in the form \(\quad k(x + m)^2 + n \quad\) where \(k\), \(m\) and \(n\) are integers. [3 marks]

(b) Solve \(\quad (x - 1)^2 - 5 = 0\) [1 mark]

Mark scheme (a)

Answer	Mark	Comments
Alternative method 1
\(2(x^2 - 8x + \ldots\ldots\) \(2(x - 4)^2 + \ldots\ldots\)	M1	oe
\(2[(x - 4)^2 - 4^2] + \ldots\ldots\) or \(2[(x - 4)^2 - 16] + \ldots\ldots\) or \(2\left[(x - 4)^2 - 16 + \dfrac{7}{2}\right]\) or \(2\left[(x - 4)^2 - \dfrac{25}{2}\right]\) or \(2(x - 4)^2 - 2 \times \dfrac{25}{2}\) or \(2(x - 4)^2 - 32 + 7\)	M1dep	oe the bracket is after the \(4^2\) and the 16 here. If they put something else inside the bracket it is incorrect unless it is equivalent to one of the fully complete versions listed
\(2(x - 4)^2 - 25\)	A1
Alternative method 2 – using identities
\(kx^2 + 2kmx + km^2 + n\)	M1	RHS oe
\(k = 2\) and \(2km = -16\) and \(km^2 + n = 7\)	M1dep	oe
\(2(x - 4)^2 - 25\)	A1

Additional guidance

Condone answer followed by \(k = 2 \quad m = -4 \quad n = -25\)	M2A1
\(k = 2 \quad m = -4 \quad n = -25\) without correct answer is penalised	M2A0

Mark scheme (b)

Answer	Mark	Comments
\((x =)\ 1 \pm \sqrt{5}\)	B1

Additional guidance

Condone roots written separately (and/or are both fine)
Do not condone \(1 + \sqrt{5}\) on its own or \(1 - \sqrt{5}\) on its own
Do not accept oe answers – must be simplified

June 2024 Paper 2 Q12

AQACurrent spec6 marksCircles in Coordinate Geometry Solving Quadratics

12 The equation of a circle is \(\quad (x - 2)^2 + (y + 3)^2 = 16\)

The equation of a line is \(\quad y = 4 - x\)

The circle and line intersect at two points, \(A\) and \(B\).

(a) Show that the \(x\)-coordinates of \(A\) and \(B\) satisfy the equation \(\quad 2x^2 - 18x + 37 = 0\) [3 marks]

(b)

For \(A\), the \(x\)-coordinate and \(y\)-coordinate are both positive.

Work out the coordinates of \(A\).

Give each coordinate to 2 decimal places. [3 marks]

Mark scheme (a)

Answer	Mark	Comments
\(x^2 - 4x + 4\) or \(y^2 + 6y + 9\) or \(49 - 14x + x^2\)	M1	oe expansion eg \(x^2 - 2x - 2x + 4\) or \(y^2 + 3y + 3y + 9\) or \(49 - 7x - 7x + x^2\) may be embedded eg in an equation
\(x^2 - 4x + 4 + 49 - 14x + x^2 = 16\) or \(x^2 - 4x + 4 + 16 - 8x + x^2 + 24 - 6x + 9 = 16\)	M1dep	oe equation in \(x\) with all relevant expansions seen eg \(x^2 - 2x - 2x + 4 + 49 - 7x - 7x + x^2 = 16\)
M2 seen and \(2x^2 - 18x + 37 = 0\) with no rearrangement errors seen	A1

Additional guidance

After M1 an expression may be simplified when forming an equation eg \(x^2 - 4x + 4 + 49 - 14x + x^2\) \(2x^2 - 18x + 53 = 16\) (equation with all relevant expansions seen) \(2x^2 - 18x + 37 = 0\) (no rearrangement errors seen)	M1 M1 A1
For M1dep the oe equation cannot be \(2x^2 - 18x + 37 = 0\)

Mark scheme (b)

Answer	Mark	Comments
\(\dfrac{--18 \pm \sqrt{(-18)^2 - 4 \times 2 \times 37}}{2 \times 2}\) or \(\dfrac{18 \pm \sqrt{28}}{4}\) or [5.8, 5.823] or [3.17, 3.2]	M1	oe calculation or value for at least one \(x\)-coordinate eg \(\dfrac{9 \pm \sqrt{7}}{2}\) or \(\dfrac{9}{2} \pm \sqrt{\dfrac{7}{4}}\) condone \(+\) for \(\pm\) condone \(-\) for \(\pm\)
\(4 - \dfrac{18 + \sqrt{28}}{4}\) or \(4 -\) their [5.8, 5.823] or [−1.823, −1.8] or \(4 - \dfrac{18 - \sqrt{28}}{4}\) or \(4 -\) their [3.17, 3.2] or [0.8, 0.83]	M1dep	oe calculation or value for at least one \(y\)-coordinate
(3.18, 0.82)	A1	both coordinates must be to 2 dp

Additional guidance

Marks may be awarded from working in (a)
Using \(-18^2\) in the discriminant is M0 unless recovered
Substitution into equation of circle is possible for the 2nd M1 eg \((y + 3)^2 = 16 - (3.17 - 2)^2\) (1st M1 for 3.17 but no calculation or value of \(y\)) \((y =)\ \sqrt{16 - (3.17 - 2)^2} - 3\)	M1 M1

June 2024 Paper 1 Q8

AQACurrent spec3 marksFactorising Solving Quadratics

8 Solve by factorising \(\quad 2x^3 - 9x^2 - 5x = 0\) [3 marks]

Mark scheme

Answer	Mark	Comments
\(x(2x^2 - 9x - 5)\) or \((2x^2 + x)(x - 5)\) or \((2x + 1)(x^2 - 5x)\) or states \(x = 0\) and \((2x^2 - 9x - 5)\)	M1
\(x(2x + 1)(x - 5)\) or states \(x = 0\) and \((2x + 1)(x - 5)\)	M1dep	do not accept \(x(2x + 1)(2x - 10)\) unless recovered by correct solutions
\((x =)\ 0,\ 5,\ -0.5\)	A1	oe but must come from factorising in M1dep

Additional guidance

\((2x + 1)(x - 5)\) if no other marks scored	SC1
\((x =)\ 0,\ 5,\ -0.5\) without working or by using factor theorem to find solutions without writing it as a product of factors or by using quadratic formula (must be all 3 correct answers)	SC1

June 2023 Paper 2 Q16

AQACurrent spec4 marksPolynomials Solving Quadratics

16 \((2x - 3)\) is a factor of \(\quad 6x^3 - 25x^2 + 28x - 6\)

Solve \(\quad 6x^3 - 25x^2 + 28x - 6 = 0\)

Give all solutions as exact values. [4 marks]

Mark scheme

Answer	Mark	Comments
\((2x - 3)(3x^2 - 8x \ldots)\) or \((2x - 3)(3x^2 \ldots + 2)\)	M1	implied by algebraic division eg \(\begin{array}{r} 3x^2 - 8x \ldots \\ 2x - 3\,\overline{\big)\,6x^3 - 25x^2 + 28x - 6} \end{array}\)
\(3x^2 - 8x + 2 \;(= 0)\) or \(3x^2 - 8x = -2\)	M1dep
\(\dfrac{- -8 \pm \sqrt{(-8)^2 - 4 \times 3 \times 2}}{2 \times 3}\) or \(\dfrac{4}{3} \pm \sqrt{\dfrac{10}{9}}\) or [2.38, 2.39] and [0.27, 0.28]	M1	oe use of formula or completing the square or factorising their 3-term quadratic eg \(\dfrac{8 \pm \sqrt{40}}{6}\) or \(\dfrac{4}{3} \pm \dfrac{\sqrt{10}}{3}\) [2.38, 2.39] and [0.27, 0.28] implies M3 \(\dfrac{8 \pm \sqrt{40}}{6}\) oe implies M3
\(\dfrac{3}{2}\) and \(\dfrac{4}{3} + \dfrac{\sqrt{10}}{3}\) and \(\dfrac{4}{3} - \dfrac{\sqrt{10}}{3}\) with no other solutions	A1	oe values eg 1.5 and \(\dfrac{8 \pm \sqrt{40}}{6}\) or \(1\dfrac{1}{2}\) and \(\dfrac{4}{3} + \sqrt{\dfrac{10}{9}}\) and \(\dfrac{4}{3} - \sqrt{\dfrac{10}{9}}\)

Additional guidance

Up to M3 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts

Up to the first two marks may be seen in a grid
eg

	\(3x^2\)	\(-8x\)	(+)2
\(2x\)	\(6x^3\)	\(-16x^2\)	\(4x\)
\(-3\)	\(-9x^2\)	\(24x\)	\(-6\)

M1M1

3rd M1 \(\;8^2\) is equivalent to \((-8)^2\)

3rd M1 Use of \(-8^2\) for \((-8)^2\) must be recovered

3rd M1 Completing the square must get as far as numerical answers which must be correct for their quadratic

3rd M1 3-term quadratic means \(ax^2 + bx + c\) with no zero coefficients

3rd M1 Allow correct substitution into quadratic formula for their 3-term quadratic even if it has no real solutions

3rd M1 can be implied by correct solutions of their 3-term quadratic truncated or rounded to 2 dp or better

A quadratic from differentiation can be used to score the 3rd M1

Non-exact solutions cannot score full marks
eg 1.5 and 2.38 and 0.28

M3A0

Non-exact solutions on answer line is A0

June 2023 Paper 1 Q15

AQACurrent spec4 marksIndices Solving Quadratics

15 Solve \(\quad \left(x^{\frac{1}{2}} - x^{\frac{3}{2}}\right)^2 = x^2 + x\) [4 marks]

Mark scheme

Answer	Mark	Comments
\(x - x^2 - x^2 + x^3 \quad (= x^2 + x)\)	M1	allow error in 1 term but must not have missing terms (could be seen in a grid) if they go straight to \(x - 2x^2 + x^3\) then M1 if they go straight to \(x - kx^2 + x^3\) where \(k\) is not 2 then this could be more than one error so M0
\(x^3 - 3x^2 = 0\) or \(x^3 = 3x^2\)	M1dep	no ft
\(x^2(x - 3)(= 0)\)	M1	factorising their equation fully and correctly (must have \(x^3\) plus/minus other term(s) in \(x^n\) where n\(\neq\)0).
(\(x =\)) 0 and (\(x =\)) 3	A1	condone \(x = 0\) \(x = 0\) \(x = 3\)

Additional guidance

First M mark could be done in root form but would still need to end up with \(x - x^2 - x^2 + x^3 \quad (= x^2 + x)\) for M1
Candidates may divide by \(x\) at some point during the method. If they do this and state that \(x = 0\) is a solution at this point, the method can be continued to a correct outcome for full marks or \(x^2 - 3x = 0\) or \(x^2 = 3x\) would be M1dep and \(x(x - 3)(= 0)\) would be the third M mark eg \(x - x^2 - x^2 + x^3 = x^2 + x\) \(1 - x - x + x^2 = x + 1\) with \(x = 0\) \(x^2 - 3x = 0\) If they do not state \(x = 0\) is a solution and continue to a correct solution then award full marks but if it’s not stated and they don’t get to the correct solution then no further marks available as soon as the division is done eg \(x - x^2 - x^2 + x^3 = x^2 + x\) \(1 - x - x + x^2 = x + 1\) (no mention of \(x = 0\) being a possible solution) \(x^2 - 3x = 0\)	M1 M1dep M1 M0dep
If no other marks awarded then \(x = 0\) or \(x = 3\) seen on answer line (or the last thing they have written)	SC1
\(x = 0\) and \(x = 3\) not from incorrect working	M3 A1

June 2022 Paper 2 Q22

AQACurrent spec4 marksIndices Solving Quadratics

22 Using powers of 2 or otherwise, work out the non-zero value of \(x\) for which

\[\left(16^x\right)^x = \dfrac{1}{2^{3x}}\]

You must show your working. [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1 Uses powers of 2
\((16^x =)\; 2^{4x}\) or \(((16^x)^x =)\; (2^4)^{x^2}\)	M1	implied by \(((16^x)^x =)\; 2^{4x^2}\) may be implied by 3rd M1
\(((16^x)^x =)\; 2^{4x^2}\)	M1dep	implied by \(2^{4x^2 + 3x}\) may be implied by 3rd M1
Correct quadratic equation or correct linear equation or correct equation involving indices with the same base	M1dep	eg \(4x^2 = -3x\) or \(4x^2 + 3x = 0\) or \(4x = -3\) or \(2^{4x^2} = 2^{-3x}\) or \(2^{4x^2 + 3x} = 2^0\) do not allow if the equation is from incorrect working do not allow if the only equation is \(x = -\dfrac{3}{4}\)
M3 and \(-\dfrac{3}{4}\)	A1	oe ignore inclusion of answer 0
Alternative method 2 Uses powers of 16
\(((16^x)^x =)\; 16^{x^2}\) or \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{\left(16^{\frac{1}{4}}\right)^{3x}}\)	M1	implied by \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{16^{\frac{3x}{4}}}\) or \(\left(\dfrac{1}{2^{3x}} =\right) 16^{-\frac{3x}{4}}\) may be implied by 3rd M1
\(((16^x)^x =)\; 16^{x^2}\) and \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{16^{\frac{3x}{4}}}\)	M1dep	oe eg \(((16^x)^x =)\; 16^{x^2}\) and \(\left(\dfrac{1}{2^{3x}} =\right) 16^{-\frac{3x}{4}}\) may be implied by 3rd M1
Correct quadratic equation or correct linear equation or correct equation involving indices with the same base	M1dep	eg \(x^2 = -\dfrac{3}{4}x\) or \(4x^2 + 3x = 0\) or \(16^{x^2} = 16^{-\frac{3x}{4}}\) do not allow if the equation is from incorrect working do not allow if the only equation is \(x = -\dfrac{3}{4}\)
M3 and \(-\dfrac{3}{4}\)	A1	oe ignore inclusion of answer 0
Alternative method 3 Uses powers of 4
\((16^x =)\; 4^{2x}\) or \(((16^x)^x =)\; (4^2)^{x^2}\) or \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{\left(4^{\frac{1}{2}}\right)^{3x}}\)	M1	implied by \(((16^x)^x =)\; 4^{2x^2}\) or \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{4^{\frac{3x}{2}}}\) or \(\left(\dfrac{1}{2^{3x}} =\right) 4^{-\frac{3}{2}x}\) may be implied by 3rd M1
\(((16^x)^x =)\; 4^{2x^2}\) and \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{4^{\frac{3x}{2}}}\)	M1dep	oe \(((16^x)^x =)\; 4^{2x^2}\) and \(\left(\dfrac{1}{2^{3x}} =\right) 4^{-\frac{3}{2}x}\) may be implied by 3rd M1
Correct quadratic equation or correct linear equation or correct equation involving indices with the same base	M1dep	eg \(2x^2 = -\dfrac{3}{2}x\) or \(4x^2 + 3x = 0\) or \(4^{2x^2} = 4^{-\frac{3}{2}x}\) do not allow if the equation is from incorrect working do not allow if the only equation is \(x = -\dfrac{3}{4}\)
M3 and \(-\dfrac{3}{4}\)	A1	oe ignore inclusion of answer 0
Alternative method 4 Takes the \(x\)th root of each side and uses powers of 2
\((16^x =)\; 2^{4x}\) or \(16^x = \left(\dfrac{1}{2^{3x}}\right)^{\frac{1}{x}}\)	M1	oe eg \(16^x = \sqrt[x]{\dfrac{1}{2^{3x}}}\) or \(16^x = \dfrac{1}{2^3}\) or \(16^x = 2^{-3}\) may be implied by 3rd M1
\(2^{4x} = \left(\dfrac{1}{2^{3x}}\right)^{\frac{1}{x}}\)	M1dep	oe eg \(2^{4x} = \dfrac{1}{2^3}\) may be implied by 3rd M1
Correct quadratic equation or correct linear equation or correct equation involving indices with the same base	M1dep	eg \(4x = -3\) or \(2^{4x} = 2^{-3}\) do not allow if the equation is from incorrect working do not allow if the only equation is \(x = -\dfrac{3}{4}\)
M3 and \(-\dfrac{3}{4}\)	A1	oe ignore inclusion of answer 0

Additional guidance

Up to M2 may be awarded for correct work with no, or incorrect answer, even if this is seen amongst multiple attempts
Allow \(2^{4 \times x \times x}\) for \(2^{4x^2}\) etc
Responses using other powers eg powers of 8 can be escalated	Escalate
Ignore simplification or conversion if correct answer seen

June 2022 Paper 1 Q16

AQACurrent spec6 marksSimultaneous Equations Solving Quadratics

16 Solve the simultaneous equations

\[\begin{aligned} x - y &= \frac{19}{4} \\ xy &= -3 \end{aligned}\]

Do not use trial and improvement.

You must show your working. [6 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
Correct substitution \(x - \dfrac{-3}{x} = \dfrac{19}{4}\) or \(x\left(x - \dfrac{19}{4}\right) = -3\)	M1	penalise no brackets unless recovered
\(4x^2 - 19x + 12 \;(= 0)\)	M1dep	oe eg \(4x^2 + 12 = 19x\) must be integer values unless going on to complete the square
\((4x + a)(x + b)\) or \((4x - 3)(4x - 16)\)	M1dep	where \(ab = 12\) or \(a + 4b = -19\)
\((4x - 3)(x - 4)\)	A1
\(x = \dfrac{3}{4}\) and 4 or \(x = \dfrac{3}{4}\) and \(y = -4\) or \(x = 4\) and \(y = -\dfrac{3}{4}\)	A1
\(y = -4\) and \(-\dfrac{3}{4}\) or \(x = 4\) and \(y = -\dfrac{3}{4}\) or \(x = \dfrac{3}{4}\) and \(y = -4\)	A1	all 4 values must be correct to gain this mark
Alternative method 2
Correct substitution \(x - \dfrac{-3}{x} = \dfrac{19}{4}\) or \(x\left(x - \dfrac{19}{4}\right) = -3\)	M1	penalise no brackets unless recovered
\(4x^2 - 19x + 12 \;(= 0)\)	M1dep	oe eg \(4x^2 + 12 = 19x\) must be integer values unless going on to complete the square
\(\dfrac{19 \pm \sqrt{19^2 - 4 \times 4 \times 12}}{2 \times 4}\)	M1dep
\(\dfrac{19 \pm \sqrt{169}}{8}\)	A1
\(x = \dfrac{3}{4}\) and 4 or \(x = \dfrac{3}{4}\) and \(y = -4\) or \(x = 4\) and \(y = -\dfrac{3}{4}\)	A1
\(y = -4\) and \(-\dfrac{3}{4}\) or \(x = 4\) and \(y = -\dfrac{3}{4}\) or \(x = \dfrac{3}{4}\) and \(y = -4\)	A1	all 4 values must be correct to gain this mark
Alternative method 3
Correct substitution \(x - \dfrac{-3}{x} = \dfrac{19}{4}\) or \(x\left(x - \dfrac{19}{4}\right) = -3\)	M1	penalise no brackets unless recovered
\(4x^2 - 19x + 12 \;(= 0)\) or \(x^2 - \dfrac{19}{4}x + 3 \;(= 0)\)	M1dep	oe eg \(4x^2 + 12 = 19x\) must be integer values unless going on to complete the square
\(4\left[\left(x - \dfrac{19}{8}\right)^2 \ldots\ldots\right] \ldots\ldots\) or \(\left[\left(x - \dfrac{19}{8}\right)^2 \ldots\ldots\right] \ldots\ldots\)	M1	oe
\(4\left(x - \dfrac{19}{8}\right)^2 - \dfrac{169}{16} = 0\) or \(\left[\left(x - \dfrac{19}{8}\right)^2\right] - \dfrac{169}{64} = 0\)	M1dep
\(x = \dfrac{3}{4}\) and 4 or \(x = \dfrac{3}{4}\) and \(y = -4\) or \(x = 4\) and \(y = -\dfrac{3}{4}\)	A1
\(y = -4\) and \(-\dfrac{3}{4}\) or \(x = 4\) and \(y = -\dfrac{3}{4}\) or \(x = \dfrac{3}{4}\) and \(y = -4\)	A1	all 4 values must be correct to gain this mark
Alternative method 4
Correct substitution \(\dfrac{-3}{y} - y = \dfrac{19}{4}\) or \(y\left(y + \dfrac{19}{4}\right) = -3\)	M1	penalise no brackets unless recovered
\(4y^2 + 19y + 12 \;(= 0)\)	M1dep	oe eg \(4y^2 + 12 = -19y\) must be integer values unless going on to complete the square
\((4y + a)(y + b)\) or \((4y + 3)(4y + 16)\)	M1dep	where \(ab = 12\) or \(a + 4b = 19\)
\((4y + 3)(y + 4)\)	A1
\(y = -\dfrac{3}{4}\) and \(-4\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\)	A1
\(x = 4\) and \(\dfrac{3}{4}\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\)	A1	all 4 values must be correct to gain this mark
Alternative method 5
Correct substitution \(\dfrac{-3}{y} - y = \dfrac{19}{4}\) or \(y\left(y + \dfrac{19}{4}\right) = -3\)	M1	penalise no brackets unless recovered
\(4y^2 + 19y + 12 \;(= 0)\)	M1dep	oe eg \(4y^2 + 12 = -19y\) must be integer values unless going on to complete the square
\(\dfrac{-19 \pm \sqrt{19^2 - 4 \times 4 \times 12}}{2 \times 4}\)	M1dep
\(-\dfrac{19 \pm \sqrt{169}}{8}\)	A1
\(y = -\dfrac{3}{4}\) and \(-4\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\)	A1
\(x = 4\) and \(\dfrac{3}{4}\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\)	A1	all 4 values must be correct to gain this mark
Alternative method 6
Correct substitution \(\dfrac{-3}{y} - y = \dfrac{19}{4}\) or \(y\left(y + \dfrac{19}{4}\right) = -3\)	M1	penalise no brackets unless recovered
\(4y^2 + 19y + 12 \;(= 0)\) or \(y^2 + \dfrac{19}{4}y + 3 \;(= 0)\)	M1dep	oe eg \(4y^2 + 12 = -19y\) must be integer values unless going on to complete the square
\(4\left[\left(y + \dfrac{19}{8}\right)^2 \ldots\ldots\right] \ldots\ldots\) or \(\left[\left(y + \dfrac{19}{8}\right)^2 \ldots\ldots\right] \ldots\ldots\)	M1	oe
\(4\left(y + \dfrac{19}{8}\right)^2 - \dfrac{169}{16} = 0\) or \(\left[\left(y + \dfrac{19}{8}\right)^2\right] - \dfrac{169}{64} = 0\)	M1dep
\(y = -\dfrac{3}{4}\) and \(-4\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\)	A1
\(x = 4\) and \(\dfrac{3}{4}\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\)	A1	all 4 values must be correct to gain this mark

Additional guidance

Correct A marks must come from correct algebra in M marks

June 2022 Paper 1 Q9

AQACurrent spec6 marksSequences Solving Quadratics

(a) A quadratic sequence starts \(\qquad -2 \qquad -1 \qquad 4 \qquad 13\)

Work out an expression for the \(n\)th term. [3 marks]

(b) A different quadratic sequence has \(n\)th term \(\quad n^2 + 10n\)

Use an algebraic method to work out how many terms in the sequence are less than 2000

Do not use trial and improvement.

You must show your working. [3 marks]

Mark scheme (a)

Answer	Mark	Comments
Alternative method 1 (grid)
\(\begin{array}{ccccc} 1 & & 5 & & 9 \\ & 4 & & 4 & \end{array}\) and \(2n^2\)	M1
\(-4 \quad -9 \quad (-14 \quad -19)\) and \(-5n\;(+\,c)\)	M1dep	subtract \(2n^2\)
\(2n^2 - 5n + 1\)	A1
Alternative method 2 (simultaneous equations)
Any 3 of: \(a + b + c = -2\) \(4a + 2b + c = -1\) \(9a + 3b + c = 4\) \(16a + 4b + c = 13\)	M1	using \(n^{\text{th}}\) term \(= an^2 + bn + c\)
\(3a + b = 1\) or \(5a + b = 5\)	M1dep	or any other equation with an unknown eliminated
\(b = -5\), \(c = 1\) so \(2n^2 - 5n + 1\)	A1
Alternative method 3 (using terms)
\(\begin{array}{ccccc} 1 & & 5 & & 9 \\ & 4 & & 4 & \end{array}\) so \(a = 2\)	M1	using \(n^{\text{th}}\) term \(= an^2 + bn + c\)
\(3a + b = 1\) and \(a = 2\) substituted in this equation	M1dep	oe
\(b = -5\), \(c = 1\) so \(2n^2 - 5n + 1\)	A1

Additional guidance

Condone other letters used eg \(2x^2 - 5x + 1\) or even \(2n^2 - 5x + 1\)
After finding \(a = 2\) they may find the 0th term to get \(c = 1\)	M2
\(2n^2 + 5n - 1\) from Alt 1 but subtracting the wrong way round	SC2

Notes

(corrected from the printed mark scheme: the fourth equation in Alternative method 2 is printed as \(16a + 9b + c = 13\); for \(n = 4\) it is \(16a + 4b + c = 13\))

Mark scheme (b)

Answer	Mark	Comments
\(n^2 + 10n - 2000 \lt 0\)	M1	the correct inequality needed for this mark and must be written in this form
\((n - 40)(n + 50)\) or \((n + 5)^2 - 25 - 2000\) or \(\dfrac{-10 \pm \sqrt{10^2 - 4 \times 1 \times -2000}}{2}\)	M1	oe inequality not needed for this mark condone + instead of \(\pm\) as the negative solution has no meaning here
39	A1

Additional guidance

Do not accept T&I
Incorrect use of inequalities can be recovered by a correct use of inequalities later in the method such as \(n \lt 40\) near the end	M2
Incorrect use of inequalities can be recovered for full marks. An answer of 39 after a method that uses an incorrect inequality or = shows inequality has been recovered	M2A1
An incorrect solution with incorrect use of inequalities can only be awarded the second M mark	M0M1A0
Correct answer not coming from correct working will not gain any marks	M0A0
For students who try to complete the square accept \((n + 5)^2 \lt 2025\) as an oe giving M2 but \((n + 5)^2 = 2025\) would only gain M0M1 unless recovered in the answer

June 2022 Paper 2 Q8

AQACurrent spec4 marksDifferentiation Solving Quadratics

8 A curve has equation \(\quad y = x^3 - 5x^2\)

At two points on the curve, the rate of change of \(y\) with respect to \(x\) is 4

(a) Work out an equation, in terms of \(x\), to represent this information.

Give your answer in the form \(\quad ax^2 + bx + c = 0 \quad\) where \(a\), \(b\) and \(c\) are integers. [2 marks]

(b) Hence, work out the two possible values of \(x\).

Give your answers to 3 significant figures. [2 marks]

Mark scheme (a)

Answer	Mark	Comments
\(3x^2\) or \(-10x\)	M1	oe eg \(3 \times x^{3 - 1}\) or \(-2 \times 5x^1\)
\(3x^2 - 10x - 4 = 0\) or \(-3x^2 + 10x + 4 = 0\)	A1	must show = 0

Additional guidance

M1 may be awarded for correct work with no, or incorrect answer, even if this is seen amongst multiple attempts
Ignore extra terms eg \(3x^2 - 10x + c\)	M1
\(3x^2 - 10x = 4\) (even if \(3x^2 - 10x - 4 = 0\) in (b))	M1A0
\(3x^2 - 10x - 4\) (even if \(3x^2 - 10x - 4 = 0\) in (b))	M1A0
\(3x^2 - 10x - 4 = 0\) seen in working with \(3x^2 - 10x - 4\) on answer line	M1A1
Condone for M1 \(\;y = 3x^2 \ldots\) etc (may still score A1 if recovered)
Answer \(y = 3x^2 - 10x - 4 = 0\)	M1A0

Mark scheme (b)

Answer	Mark	Comments
\(\dfrac{- -10 \pm \sqrt{(-10)^2 - 4 \times 3 \times -4}}{2 \times 3}\) or \(\dfrac{10 \pm \sqrt{148}}{6}\) or \(\dfrac{5}{3} \pm \sqrt{\dfrac{37}{9}}\) or two correct solutions with at least one not to 3 sf	M1	oe eg \(\dfrac{5 \pm \sqrt{37}}{3}\) correct attempt to solve their \(ax^2 + bx + c\;(= 0)\) from (a) \(a\), \(b\) and \(c\) all non-zero eg 3.69(4…) and \(-0.36(09\ldots)\) or 3.7 and \(-0.36(09\ldots)\)
3.69 and \(-0.361\)	A1ft	correct or ft any answers that have at least 4 sf must be rounded to 3 sf at least one answer must have at least 4 sf

Additional guidance

\(-10^2\) used for \((-10)^2\) is M0 unless recovered
\(10^2\) is equivalent to \((-10)^2\)
Not using \(\pm\) is M0 unless recovered
A short dividing line or a short square root symbol is M0 unless recovered
√((–10)\(^2\) – 4 × 3 × –4) is correct for \(\sqrt{(-10)^2 - 4 \times 3 \times -4}\)
Correct factorisation of their \(ax^2 + bx + c\;(= 0)\) from (a) scores at least M1
(a) \(3x^2 - 10x + 4 = 0 \qquad\) (b) \(\dfrac{- -10 \pm \sqrt{(-10)^2 - 4 \times 3 \times 4}}{2 \times 3}\) 2.87 and 0.465	M1A1ft
(a) \(3x^2 - 10x = 4 \qquad\) (b) up to 2 marks can be scored if using \(3x^2 - 10x - 4 = 0\)
(a) \(3x^2 - 10x - 8 \qquad\) (b) up to 2 marks can be scored if using \(3x^2 - 10x - 8 = 0\)
One solution correct does not imply M1
Both solutions seen in working but only one on answer line	M1A0
3.69 and \(-0.361\) in working with \(-3.69\) and 0.361 on answer line	M1A0

Specimen 2020 Paper 2 Q17

AQACurrent spec4 marksIndices Solving Quadratics

17 By multiplying both sides of the equation by \(x^{\frac{1}{2}}\)

Solve \(\quad 2x^{\frac{3}{2}} - 3x^{\frac{1}{2}} = 7x^{-\frac{1}{2}} \qquad\) for \(\quad x \gt 0\)

Give your answer to 3 significant figures.

You must show your working. [4 marks]

Mark scheme

Answer	Mark	Comments
\(2x^2 - 3x = 7\)	M1	at least two terms correct
\(2x^2 - 3x - 7\ (= 0)\)	A1	oe 3-term quadratic equation
\(\dfrac{--3 \pm \sqrt{(-3)^2 - 4 \times 2 \times -7}}{2 \times 2}\) or \(\dfrac{3}{4} \pm \sqrt{\dfrac{65}{16}}\)	M1	oe correct attempt to solve their 3-term quadratic equation
2.77	A1	2.77 and \(-1.27\) is A0

Specimen 2020 Paper 2 Q11

AQACurrent spec4 marksSequences Solving Quadratics

11 For sequence A, \(\quad n\)th term \(= \dfrac{n}{14n + 30}\)

For sequence B, \(\quad n\)th term \(= \dfrac{2}{n}\)

The \(k\)th term of sequence A equals the \(k\)th term of sequence B.

Work out the value of \(k\).

You must show your working. [4 marks]

Mark scheme

Answer	Mark	Comments
\(k^2 = 2(14k + 30)\)	M1	oe correct equation with fractions eliminated
\(k^2 - 28k - 60\ (= 0)\)	M1dep	oe equation
\((k + 2)(k - 30)\ (= 0)\) or \(\dfrac{--28 \pm \sqrt{(-28)^2 - 4 \times 1 \times -60}}{2 \times 1}\) or \(14 \pm \sqrt{256}\)	M1	oe correct attempt to solve their 3-term quadratic equation
30	A1	30 and \(-2\) is A0

Specimen 2020 Paper 2 Q5

AQACurrent spec2 marksDrawing & Using Graphs Solving Quadratics

5 Here is a sketch of \(\quad y = \mathrm{f}(x) \quad\) where \(\mathrm{f}(x)\) is a quadratic function.

The graph

intersects the \(x\)-axis at \(A\ (-1, 0)\) and \(B\)
has a maximum point at \(\quad (0.5, 6)\)

Sketch of an n-shaped parabola crossing the x-axis at A (left of O) and B (right of O)

Not drawn accurately

(a) Work out the coordinates of \(B\). [1 mark]

(b) The equation \(\ \mathrm{f}(x) = k\ \) has exactly one solution.

Write down the value of \(k\). [1 mark]

Answer	Mark	Comments
(2, 0)	B1

Answer	Mark	Comments
6	B1

June 2019 Paper 2 Q18

AQAOld spec3 marksSolving Quadratics

18 Solve \(\quad 4(x - 5)^2 = k^2 \quad\) where \(k\) is a constant.

Give your answers in their simplest form in terms of \(k\). [3 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(x - 5 = \dfrac{k}{2}\) or \(x - 5 = -\dfrac{k}{2}\) or \(2(x - 5) = k\) or \(2x - 10 = k\) or \(2(x - 5) = -k\) or \(2x - 10 = -k\)	M1	oe linear equation eg \(x - 5 = \sqrt{\dfrac{k^2}{4}}\) or \(x = \dfrac{k}{2} + 5\) or \(\sqrt{4}\,(x - 5) = \sqrt{k^2}\)
\(x - 5 = \dfrac{k}{2}\) and \(x - 5 = -\dfrac{k}{2}\) or \(2(x - 5) = k\) and \(2(x - 5) = -k\) or \(2x - 10 = k\) and \(2x - 10 = -k\)	A1	oe eg \(x - 5 = \pm\dfrac{k}{2}\) square root(s) must be processed implied by final A1
\(\dfrac{k}{2} + 5\) and \(-\dfrac{k}{2} + 5\)	A1	oe simplest form eg \(\dfrac{10 + k}{2}\) and \(\dfrac{10 - k}{2}\) or \(\dfrac{k + 10}{2}\) and \(\dfrac{k - 10}{-2}\) or \(5 \pm 0.5k\)
Alternative method 2
\(4x^2 - 40x + 100 - k^2\ (= 0)\)	M1	expands and collects terms
\(\dfrac{-40 \pm \sqrt{(-40)^2 - 4 \times 4 \times (100 - k^2)}}{2 \times 4}\)	A1	oe eg \(\dfrac{40 \pm \sqrt{16k^2}}{8}\) or \(\dfrac{40 \pm 4k}{8}\) implied by final A1
\(\dfrac{k}{2} + 5\) and \(-\dfrac{k}{2} + 5\)	A1	oe simplest form eg \(\dfrac{10 + k}{2}\) and \(\dfrac{10 - k}{2}\) or \(\dfrac{k + 10}{2}\) and \(\dfrac{k - 10}{-2}\) or \(5 \pm 0.5k\)

Additional guidance

Allow recovery of missing brackets

June 2019 Paper 2 Q6

AQAOld spec3 marksInequalities Solving Quadratics

6 Here is a sketch of the curve \(\quad y = (2x + 3)(x - 2)\)

The curve intersects the \(x\)-axis at \(A\) and \(B\).

U-shaped parabola crossing the x-axis at A (left of O) and B (right of O). Not drawn accurately

(a) Complete the coordinates of \(A\) and \(B\). [2 marks]

\(A\ (\ \_\_\_\_\_\ ,\ 0\ ) \qquad B\ (\ \_\_\_\_\_\ ,\ 0\ )\)

(b) Write down the range of values for \(x\) for which \(\quad (2x + 3)(x - 2) \lt 0\) [1 mark]

Mark scheme (a)

Answer	Mark	Comments
\(A\left(-\dfrac{3}{2},\ 0\right)\) and \(B\,(2,\ 0)\)	B2	oe B1 \(A\left(-\dfrac{3}{2},\ 0\right)\) oe or \(B\,(2,\ 0)\) SC1 \(A\,(2,\ 0)\) and \(B\left(-\dfrac{3}{2},\ 0\right)\) oe

Additional guidance

Ignore the diagram

Mark scheme (b)

Answer	Mark	Comments
\(-\dfrac{3}{2} \lt x \lt 2\) or \(2 \gt x \gt -\dfrac{3}{2}\)	B1ft	oe correct or ft their values from (a) must be a single inequality in \(x\)

Additional guidance

\(-\dfrac{3}{2} \leqslant x \lt 2\)	B0
\(-\dfrac{3}{2} \gt x \lt 2\)	B0
\(-\dfrac{3}{2} \lt x\) and \(x \lt 2\)	B0
their (a) \(A\,(-2,\ 0)\) and \(B\left(\dfrac{3}{2},\ 0\right)\) (B0 in (a)) (b) \(-2 \lt x \lt \dfrac{3}{2}\)	B1ft
their (a) \(A\,(2,\ 0)\) and \(B\left(-\dfrac{3}{2},\ 0\right)\) (SC1 in (a)) (b) \(2 \lt x \lt -\dfrac{3}{2}\)	B0ft
their (a) \(A\,(-3,\ 0)\) and \(B\,(2,\ 0)\) (B1 in (a)) (b) \(2 \lt x \lt -3\)	B0ft
their (a) \(A\,(4,\ 0)\) and \(B\,(-2,\ 0)\) (B0 in (a)) (b) \(-2 \lt x \lt 4\)	B1ft
Only one value in (a) can only score in (b) for \(-\dfrac{3}{2} \lt x \lt 2\) or \(2 \gt x \gt -\dfrac{3}{2}\)

June 2018 Paper 1 Q15

AQAOld spec4 marksDrawing & Using Graphs Solving Quadratics

15 Here is the graph of \(\quad y = 3x - x^2 \quad\) for values of \(x\) from \(-1\) to 4

Graph of y = 3x − x² on a 2 mm grid, x from −1 to 4 and y from −4 to 3; the curve passes through (0, 0) and (3, 0) with maximum near (1.5, 2.25)

By drawing a suitable linear graph on the grid, work out approximate solutions to

\[x^2 - 4x + 2 = 0\]

[4 marks]

Mark scheme

Answer	Mark	Comments
\(2 - x\) or \(x - 2\)	M1	do not award M1 if you see evidence of incorrect method for finding a linear expression
\(y = 2 - x\) accurately drawn	M1
3.4	A1	accept 3.3 to 3.5
0.6	A1	accept 0.5 to 0.7

Additional guidance

For the first M1, start by looking for evidence of a correct method.
eg \(\ x^2 - 4x + 2 + 3x - x^2 = -x + 2\)
or
\(x^2 - 4x + 2 = 0 \ \rightarrow\ x^2 - 3x - x + 2 = 0 \ \rightarrow\ -x + 2 = 3x - x^2\)

Attempts to solve \(x^2 - 4x + 2 = 0\) by using the quadratic formula or by completing the square or by drawing a new quadratic graph (for \(y = x^2 - 4x + 2\)) score 0 marks
You might see work which uses the quadratic formula or completing the square which leads to answers of \(2 \pm \sqrt{2}\) … and if this follows working using a correct method to find the linear graph, it can be ignored (they could be using it as a check on their answers obtained graphically), but if it looks like it is their main method, then award 0 marks, as stated above..
Ignore any \(y\) coordinates that might accompany the final \(x\) values.

June 2018 Paper 1 Q9

AQAOld spec4 marksFunctions Solving Quadratics

9 \(\mathrm{f}(x) = 14 - x^2 \quad\) for all real values of \(x\).

Solve \(\quad \mathrm{f}(2x) = 5\)

You must show your working. [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(14 - (2x)^2\) or \(14 - 4x^2\)	M1	or \(14 - (2x)^2 = 5\) or \(14 - 4x^2 = 5\)
\(14 - 5 = (2x)^2\) or \(9 = 4x^2\) or \(9 - 4x^2 = 0\) or \(4x^2 - 9 = 0\) or \((2x + 3)(2x - 3) = 0\)	M1dep
\((x =)\ \dfrac{3}{2}\) or 1.5	A1
\((x =)\ -\dfrac{3}{2}\) or \(-1.5\)	A1
Alternative method 2
\(14 - x^2 = 5\) and \(x = \pm 3\)	M1
\(2x = \pm 3\)	M1dep
\((x =)\ \dfrac{3}{2}\) or 1.5	A1
\((x =)\ -\dfrac{3}{2}\) or \(-1.5\)	A1

Additional guidance

A final answer of \(\sqrt{\dfrac{9}{4}}\) scores M1 M1 A0 A0

A final answer of \(\pm\sqrt{\dfrac{9}{4}}\) scores M1 M1 A1 A0

June 2018 Paper 2 Q6

AQAOld spec7 marksFunctions Solving Quadratics

6 \(\mathrm{f}(x) = x^2 - 7 \qquad\) for all values of \(x\)

\(\mathrm{g}(x) = 1 - 3x \qquad\) for \(\ -4 \leqslant x \leqslant 4\)

(a) Work out the range of \(\mathrm{f}(x)\).

Give your answer as an inequality. [1 mark]

(b) Work out the range of \(\mathrm{g}(x)\).

Give your answer as an inequality. [2 marks]

You must show your working.
Give your answers to 3 decimal places. [4 marks]

Mark scheme (a)

Answer	Mark	Comments
\(\mathrm{f}(x) \geqslant -7\) or \(-7 \leqslant \mathrm{f}(x)\)	B1

Additional guidance

\(\mathrm{f}(x)\) may be replaced by \(y\) or f or f\(x\) or g(\(x\)) or g or g\(x\) or \(x^2 - 7\)
\(x \geqslant -7\)	B0
\(\geqslant -7\)	B0
Condone \(-7 \leqslant \mathrm{f}(x) \lt \infty\) or \(-7 \leqslant \mathrm{f}(x) \leqslant \infty\) or \(-7 \leqslant \mathrm{f}(x) \lt\) or \(-7 \leqslant \mathrm{f}(x) \leqslant\)	B1
\([-7,\ \infty)\) or \([-7,\ \infty]\)	B0

Mark scheme (b)

Answer	Mark	Comments
\(-11 \leqslant \mathrm{g}(x) \leqslant 13\) or \(13 \geqslant \mathrm{g}(x) \geqslant -11\)	B2	B1 \(\mathrm{g}(x) \geqslant -11\) or \(\mathrm{g}(x) \leqslant 13\) on their own or embedded within an inequality or \(-11 \lt \mathrm{g}(x) \lt 13\) or \([-11,\ 13]\) or \(-11 \leqslant x \leqslant 13\)

Additional guidance

\(\mathrm{g}(x)\) may be replaced by \(y\) or g or g\(x\) or f(\(x\)) or f or f\(x\) or \(1 - 3x\) in B2 or B1 responses
\(\mathrm{g}(x) \geqslant -11 \quad \mathrm{g}(x) \leqslant 13\)	B1
\(-11\) to 13 inclusive (‘inclusive’ must be seen) Do not allow if 24 also seen	B1
B1 may be seen with an incorrect inequality eg1 \(\ -11 \lt \mathrm{g}(x) \leqslant 13\) eg2 \(\ -11 \leqslant \mathrm{g}(x) \lt 13\) eg3 \(\ 0 \lt \mathrm{g}(x) \leqslant 13\) eg4 \(\ 13 \leqslant \mathrm{g}(x) \geqslant -11\)	B1 B1 B1 B1
\([-11,\ 13)\) or \((-11,\ 13]\) or \((-11,\ 13)\)	B0
\(-11 \lt x \leqslant 13\) or \(-11 \leqslant x \lt 13\) or \(-11 \lt x \lt 13\)	B0
\(\{-11,\ -10,\ -9,\ \ldots\ldots\ 0,\ 1,\ 2,\ 3,\ \ldots\ldots,\ 12,\ 13\}\)	B0

Mark scheme (c)

Answer	Mark	Comments
\(2x^2 - 14\)	M1
\(2x^2 + 3x - 15\ (= 0)\) or \(-2x^2 - 3x + 15\ (= 0)\) or \(2x^2 + 3x = 15\) or \(-2x^2 - 3x = -15\)	A1
\(\dfrac{-3 \pm \sqrt{3^2 - 4 \times 2 \times -15}}{2 \times 2}\) or \(\dfrac{-3 \pm \sqrt{9 + 120}}{4}\) or \(\dfrac{-3 \pm \sqrt{129}}{4}\)	M1	oe eg \(-\dfrac{3}{4} \pm \sqrt{\dfrac{15}{2} + \left(\dfrac{3}{4}\right)^2}\) correct method to solve their 3-term quadratic implied by correct solutions to their 3-term quadratic to at least 2 dp
2.089 \(\quad -3.589\)	A1ft	correct or ft M1A0M1 or M0A0M1 must both be rounded to 3 decimal places

Additional guidance

2nd M1 Allow correct factorisation of their 3-term quadratic if it does factorise
2nd M1 Allow correct use of formula even if discriminant is negative
Two ‘correct’ solutions to at least 2 decimal places implies M1A1M1 eg 2.09 and \(-3.59\)	M1A1M1A0
2.089 and \(-3.589\) in working but only one on answer line	M1A1M1A0
Answers only 2.089 \(\quad -3.589\)	M1A1M1A1
Answer only 2.089	Zero
Answer only \(-3.589\)	Zero
\(2x^2 - 7\) from incorrect expansion leading to 1.386 \(\quad -2.886\)	M0A0M1A1ft
\(x^2 - 14\) from incorrect expansion leading to 2.653 \(\quad -5.653\)	M0A0M1A1ft
\(2x^2 - 14\) and \(2x^2 + 3x - 13\ (= 0)\) Answers 1.908 \(\quad -3.408\)	M1A0 M1A1ft

June 2018 Paper 2 Q3

AQAOld spec4 marksAlgebraic Fractions Solving Quadratics

3 \(p = \dfrac{m + 2}{m^2 + 1}\)

(a) Work out the value of \(p\) when \(\ m = -5.5\) [1 mark]

(b) Work out the values of \(m\) when \(\ p = 2\) [3 marks]

Mark scheme (a)

Answer	Mark	Comments
\(-0.112\) or \(-\dfrac{14}{125}\)	B1	oe fraction

Additional guidance

Ignore incorrect conversion between fraction and decimal if correct value seen
Ignore rounding or truncation after correct value seen
Answer \(-\dfrac{3.5}{31.25}\)	B0

Mark scheme (b)

Answer	Mark	Comments
\(2(m^2 + 1) = m + 2\) or \(2m^2 + 2 = m + 2\) or \(2m^2 = m\) or \(2m^2 - m = 0\)	M1	oe equation without fractions
\(m(2m - 1)\ (= 0)\) or \(m(1 - 2m)\ (= 0)\) or \(\dfrac{--1 \pm \sqrt{(-1)^2 - 4 \times 2 \times 0}}{2 \times 2}\)	M1dep	oe eg \(\dfrac{1}{4} \pm \sqrt{\dfrac{1}{16}}\) may be implied by both correct solutions
0.5 \(\quad\) 0 or \(\dfrac{1}{2} \quad 0\)	A1	oe

Additional guidance

0.5 \(\quad\) 0 in working but only one of these on answer line	M2A0
Equation left in terms of \(p\) and \(m\)	Zero
Answers only of 0.5 \(\quad\) 0	M2A1
Answer only of 0.5	Zero
Answer only of 0	Zero
If using formula with an error seen the maximum mark is M1M0A0 eg \(\ 2m^2 - m = 0 \quad \dfrac{--1 \pm \sqrt{(-1)^2 - 4 \times 1 \times 0}}{2 \times 2}\) Answers 0.5 \(\quad\) 0	M1M0A0

June 2017 Paper 1 Q9

AQAOld spec5 marksInequalities Solving Quadratics

9 Here is a rectangle.

Rectangle with length (2x − 3) cm and width (x + 1) cm

Not drawn accurately

(a) Show that the area of the rectangle is \(\quad 2x^2 - x - 3\) cm\(^2\) [1 mark]

(b) The area of the rectangle is greater than 7 cm\(^2\)

Work out the range of possible values of \(x\).

Give your answer as an inequality. [4 marks]

Mark scheme (b)

Answer	Mark	Comments
\(2x^2 - x - 10\) (\(\gt 0\))	M1	oe must have three terms eg \(2x^2 \gt x + 10\) Condone \(2x^2 - x - 10\) (\(= 0\)) oe eg \(2x^2 - x = 10\)
correct factors \(\quad (2x - 5)(x + 2)\)	M1dep	For use of quadratic formula ... condone one numerical or sign error
Sight of 2.5 and \(-2\)	A1
\(x \gt 2.5\)	A1	oe Must have seen \(x = 2.5\) and \(x = -2\) and reject the negative solution

Additional guidance

This is a quadratic inequality so we need to see an attempt at finding the two critical values then making a decision as to the correct solution

SC1 for \(x \gt 2.5\) with no working SC1 for \(x \gt 2.5\) from T&I

If T&I done such that both critical values (2.5 and \(-2\)) have been identified and tested (eg they give 7 when substituted in the expression \(2x^2 - x - 3\)) then it is possible to score 3 marks or 4 marks

June 2016 Paper 2 Q21

AQAOld spec6 marksAlgebraic Fractions Solving Quadratics

21 Solve \(\quad \dfrac{3}{x - 2} + \dfrac{2}{x - 1} = 5\)

Do not use trial and improvement.

Write your solutions to 3 significant figures. [6 marks]

Mark scheme

Answer	Mark	Comments
\(3(x - 1)\) or \(3x - 3\) or \(2(x - 2)\) or \(2x - 4\)	M1
\(3x - 3\) and \(2x - 4\) or \(5x - 7\)	M1	Implies M1 M1
\(5(x - 1)(x - 2)\) or \(5(x^2 - 2x - x + 2)\) or \(5x^2 - 15x + 10\) or \((x - 1)(x - 2)\) expanded and multiplied by 5	M1	oe Allow one error in four term expansion of \(5(x - 1)(x - 2)\) Implied by \(5(x^2 - 3x + k)\) or \(5(ax^2 - 3x + 2)\)
\(5x^2 - 20x + 17\ (= 0)\)	M1dep	dep on 3rd M1 oe 3-term quadratic equation eg \(5x^2 - 20x = -17\) Correctly collects terms in their expansion
\(\dfrac{--20 \pm \sqrt{(-20)^2 - 4 \times 5 \times 17}}{2 \times 5}\) or \(\dfrac{10 \pm \sqrt{15}}{5}\) or \((x - 2)^2 - 4 = -\dfrac{17}{5}\) or \(5[(x - 2)^2 - 4] = -17\)	M1	oe Correct use of quadratic formula for their 3-term quadratic eg \((-20)^2\) can be \(20^2\) or correct factorisation of their 3-term quadratic or attempt to complete the square for their 3-term quadratic Must be correct up to form \((x - a)^2 + b = c\) or \(k[(x - d)^2 + e] = f\)
1.23 and 2.77	A1	Must be 3 significant figures

Additional guidance

For A1, the word ‘and’ is not needed eg 1.23, 2.77 (with method seen)	M5 A1
Brackets may be recovered throughout
5th M1 may be implied by solutions of their quadratic equation seen
M0 M0 M0 M0 M1 A0 is possible if they have a 3-term quadratic equation
Answers only	Zero

June 2016 Paper 2 Q13

AQAOld spec4 marksDrawing & Using Graphs Solving Quadratics

13 Here is a sketch of \(\quad y = a + bx - 2x^2 \quad\) where \(a\) and \(b\) are constants.

The graph intersects the \(x\)-axis at \((-1, 0)\) and \(\left(\dfrac{7}{2}, 0\right)\) and the \(y\)-axis at point \(P\).

Sketch of an n-shaped parabola crossing the x-axis at −1 and 7/2 and the y-axis at P

Not drawn accurately

Work out the coordinates of point \(P\).

You must show your working. [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(a - b - 2 = 0\) or \(a + \dfrac{7}{2}b - \dfrac{49}{2} = 0\)	M1	oe equation Allow an unsimplified equation eg \(a + b \times (-1) - 2(-1)^2 = 0\) Missing brackets can be recovered
\(a - b - 2 = 0\) and \(a + \dfrac{7}{2}b - \dfrac{49}{2} = 0\)	M1	oe two equations Allow unsimplified equations
\(\dfrac{7}{2}a + a = 7 + \dfrac{49}{2}\) or \(\dfrac{7}{2}b - -b = \dfrac{49}{2} - 2\) and \(a\) − their \(b - 2 = 0\)	M1dep	oe dep on first M1 Correct method to form an equation in \(a\) or correct method to form an equation in \(b\) and substitutes to form an equation in \(a\) Their two equations must both contain \(a\) and \(b\)
\((0, 7)\)	A1	SC4 Answer \((0, 7)\) from \((2x - 7)(1 + x)\) or \(2x^2 + 2x - 7x - 7\) or \(2x^2 - 5x - 7\)
Alternative method 2
\((7 - 2x)\) or \((1 + x)\) or \((2x - 7)\) or \((-1 - x)\) or \((x - 3.5)\) or \((3.5 - x)\) or \(2x^2 + 2x - 7x - 7\) or \(2x^2 - 5x - 7\)	M1	oe Brackets not needed
\((7 - 2x)(1 + x)\) or \((2x - 7)(-1 - x)\) or \(7 + 7x - 2x - 2x^2\) or \(7 + 5x - 2x^2\)	M1	oe eg \(-(2x^2 - 5x - 7)\) \(y =\) not needed Expansion not needed
Substitutes \(x = 0\) in their quadratic or selects the constant term from their quadratic	M1dep	dep on first M1 Expansion not needed for their quadratic but must be correct if attempted May be implied by the final answer
\((0, 7)\)	A1	SC4 Answer \((0, 7)\) from \((2x - 7)(1 + x)\) or \(2x^2 + 2x - 7x - 7\) or \(2x^2 - 5x - 7\)
Alternative method 3
\(b - 2 \times 2x\) or \(b - 4x\)	M1	Differentiates correctly
\(b - 2 \times 2 \times 1.25 = 0\) or \(b - 4 \times 1.25 = 0\) or \(b = 5\)	M1
\(a\) + their \(b \times (-1) - 2 \times (-1)^2 = 0\) or \(a\) + their \(b \times \left(\dfrac{7}{2}\right) - 2 \times \left(\dfrac{7}{2}\right)^2 = 0\)	M1dep	oe dep on first M1 Must have substituted a value into \(b - 4x\) and equated to 0 Missing brackets can be recovered
\((0, 7)\)	A1	SC4 Answer \((0, 7)\) from \((2x - 7)(1 + x)\) or \(2x^2 + 2x - 7x - 7\) or \(2x^2 - 5x - 7\)
Alternative method 4
\(-2\left\{\left(x - \dfrac{b}{4}\right)^2 \ldots\ldots\right\}\)	M1	oe completed square
\(\dfrac{b}{4} = 1.25\) or \(b = 5\)	M1
\(a\) + their \(b \times (-1) - 2 \times (-1)^2 = 0\) or \(a\) + their \(b \times \left(\dfrac{7}{2}\right) - 2 \times \left(\dfrac{7}{2}\right)^2 = 0\)	M1dep	oe dep on first M1 Must have equated \(\dfrac{b}{4}\) to a value Missing brackets can be recovered
\((0, 7)\)	A1	SC4 Answer \((0, 7)\) from \((2x - 7)(1 + x)\) or \(2x^2 + 2x - 7x - 7\) or \(2x^2 - 5x - 7\)

Additional guidance

Answer \((7, 0)\)	A0
Answer only \((0, 7)\)	M3 A1
Alt 1 3rd M1 Follow correct method rules as in Q4 Alt 1 2nd M1
Answer \((0, 7)\) with working may not score full marks Check the working each time eg (Alt 2) \(2x^2 - 3x + 7\) (no method seen) Answer \((0, 7)\)	Zero

June 2016 Paper 1 Q12

AQAOld spec4 marksCompleting the Square Solving Quadratics

12 You are given that \(\qquad x^2 + 6x + 2 \equiv (x + h)^2 + k\)

(a) Work out the values of \(h\) and \(k\). [2 marks]

(b) Write down the coordinates of the minimum point on the curve \(\quad y = x^2 + 6x + 2\) [1 mark]

Give your answers in the form \(\ a \pm \sqrt{b}\) [1 mark]

Mark scheme (a)

Answer	Mark	Comments
Alternative method 1
\((x + 3)^2 - 9\ (+ 2)\)	M1
\(h = 3\) and \(k = -7\)	A1
Alternative method 2
\(x^2 + 2hx + h^2\ (+ k)\) or \(2hx = 6x\) or \(2h = 6\) or \(h^2 + k = 2\)	M1
\(h = 3\) and \(k = -7\)	A1

Additional guidance

\(h = 3\) implies M1

Mark scheme (b)

Answer	Mark	Comments
\((-3, -7)\)	B1 ft	ft their \(h\) and \(k\) from part (a) only if \(h \neq 0\) and \(k \neq 0\)

Additional guidance

for their \(h\) and \(k\), the minimum point is \((-h, k)\)

Mark scheme (c)

Answer	Mark	Comments
\(-3 \pm \sqrt{7}\)	B1 ft	ft their \(h\) and \(k\) from part (a) only if \(h \neq 0\) and \(k \neq 0\)

Additional guidance

For their \(h\) and \(k\), the solutions are \(-h \pm \sqrt{(-k)}\)

If their \(k\) is \(\gt 0\) then \(\sqrt{(-k)}\) will be \(\sqrt{}\) of a negative number ... condone

Any use of the quadratic formula must be completely correct

June 2016 Paper 1 Q7

AQAOld spec3 marksInequalities Solving Quadratics

7 Work out the integer values of \(x\) for which \(\qquad x^2 - 20x + 96 \lt 0\) [3 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\((x \pm a)(x \pm b)\)	M1	\(ab = 96\) or \(a + b = -20\)
\((x - 8)(x - 12)\) or (\(x\) =) 8 and (\(x\) =) 12	A1
9, 10 and 11	A1	A0 if extra values seen
Alternative method 2
\((x - 10)^2 - 100\ (+ 96)\ (\lt 0)\)	M1	oe eg \(\ (x - 10)^2 - 4\ (\lt 0)\)
\(8 \lt x \lt 12\) or (\(x\) =) 8 and (\(x\) =) 12	A1
9, 10 and 11	A1	A0 if extra values seen
Alternative method 3
\(\dfrac{20 \pm \sqrt{\{(20)^2 - 4 \times 1 \times 96\}}}{2}\) or \(\dfrac{20 \pm \sqrt{\{(-20)^2 - 4 \times 1 \times 96\}}}{2}\)	M1	accept \((20)^2\) or \((-20)^2\) for \(b^2\) in the discriminant
8 and 12	A1
9, 10 and 11	A1	A0 if extra values seen

Additional guidance

9, 10 and 11 using Trial and Improvement - all correct is 3 marks, otherwise 0 marks.

No working ... treat as Trial and Improvement

For alt 3 ... substitution in the formula must be correct