24 questions working through all three methods of solving a quadratic, plus roots and turning points from the graph.

📈 Quadratics: factorise, complete the square, formula

Maths · GCSE Higher · 24 questions · ~45 min

MathsUKwww.geekhero.co.uk

Almost every Higher paper asks you to solve a quadratic, and the mark is often lost not on the algebra but on choosing the wrong method for the quadratic in front of you. This sheet takes the three methods in turn and then mixes them. It starts with factorising, including ax² + bx + c and the difference of two squares; moves to completing the square, and to reading the turning point straight out of the completed form; then to the quadratic formula, where the marks go missing on −b when b is already negative and on dividing the whole numerator by 2a. The last section gives you quadratics with no instruction, so you have to decide for yourself. Roots, intercepts and turning points from the graph are included too, because the graphical and algebraic halves of this topic are examined together.

1.Solve 4x² − 9 = 0.
(a)x = 3/2 or x = −3/2
(b)x = 3/2 only
(c)x = 3 or x = −3
(d)x = 9/4 or x = −9/4
2.Write y = x² + 12x + 40 in the form (x + a)² + b, and hence write down the minimum value of y.
y = x² + 12x + 40
(a)−6
(b)36
(c)4
(d)40
3.A rectangle has length (2x + 5) cm and width (x − 2) cm. Work out an expression, in terms of x, for the perimeter of the rectangle. Give your answer in its simplest form.
(a)4x² + 2x − 20
(b)6x + 6
(c)6x + 14
(d)3x + 3
4.The graph of y = x² − 5x + 6 crosses the x-axis at two points. By factorising, work out the x-coordinates of these two points.
y = x² − 5x + 6
(a)x = −1 and x = −6
(b)x = 2 and x = −3
(c)x = 2 and x = 3
(d)x = −2 and x = −3
5.Solve x² − 8x + 3 = 0 by completing the square. Give your answers in surd form.
(a)x = 4 ± √13
(b)x = −4 ± √13
(c)x = 8 ± √13
(d)x = 4 ± √19
6.Simplify fully: (x + 2)/(3x) × 6x²/(x² − 4)
(a)2/(x − 2)
(b)2x/(x + 2)
(c)2x²/(x − 2)
(d)2x/(x − 2)
7.A ball is thrown in the air. Its height, h metres, above the ground after t seconds is given in this table: when t = 0, h = 0; when t = 1, h = 15; when t = 2, h = 20; when t = 3, h = 15; when t = 4, h = 0. Use the table to find the two times, in seconds, at which the ball is at ground level.
(a)t = 0 or t = 4
(b)t = 2
(c)t = 0
(d)t = 1 or t = 3
8.Solve 2x² − 7x + 3 = 0.
(a)x = 6 or x = 1
(b)x = 3/2 or x = 1
(c)x = 3 or x = 1/2
(d)x = −3 or x = −1/2
9.The curve y = −x² + 6x − 5 has a maximum point. Use completing the square to find its coordinates.
y = −x² + 6x − 5
(a)x = 3, y = −4
(b)x = 3, y = 4
(c)x = 6, y = 31
(d)x = −3, y = 4
10.Simplify (5x² + 3x − 2) − (2x² − x + 5)
(a)3x² + 4x + 3
(b)3x² + 2x + 3
(c)7x² + 2x + 3
(d)3x² + 4x − 7
11.Simplify (x² − 9) ÷ (x² + 5x + 6).
(a)(x + 3)/(x + 2)
(b)(x − 3)/(x + 2)
(c)x − 3
(d)−9/(5x + 6)
12.A quadratic graph y = ax² + bx + c has its turning point on the y-axis. Which statement about its roots must be true?
(a)The two roots must both be positive.
(b)The graph has no roots at all.
(c)The graph must touch the x-axis at exactly one point.
(d)If the graph has two real roots, they are equal and opposite in value, so they sum to zero.
13.Factorise fully 12x² + 18x.
(a)2x(6x + 9)
(b)6(2x² + 3x)
(c)6x(2x + 3)
(d)3x(4x + 6)
14.One solution of the equation x² − (k + 1)x + k = 0 is x = 3. Work out the value of k.
(a)k = 3
(b)k = −3
(c)k = 4
(d)k = 1.5
15.Subtract, giving your answer as a single fraction in its simplest form: 4/(x − 1) − 2/(x + 3)
(a)(2x + 11)/((x − 1)(x + 3))
(b)(2x + 14)/((x − 1)(x + 3))
(c)2/((x − 1)(x + 3))
(d)(2x − 10)/((x − 1)(x + 3))
16.Factorise 3x² + 10x − 8.
(a)(3x − 4)(x + 2)
(b)(3x + 2)(x − 4)
(c)(x − 2)(3x + 4)
(d)(3x − 2)(x + 4)
17.The equation x² − 6x + k = 0 has exactly one solution. Work out the value of k.
(a)k = −9
(b)k = 9
(c)k = 3
(d)k = 36
18.Factorise x² − 64.
(a)(x − 4)(x + 16)
(b)(x − 8)(x − 8)
(c)(x + 8)(x + 8)
(d)(x − 8)(x + 8)
19.A rectangular lawn is 4 m longer than it is wide. Its area is 96 m². Work out the length of the lawn.
(a)8 m
(b)24 m
(c)16 m
(d)12 m
20.The graph of y = x² − 7x + 2 crosses the x-axis at two points. One root, read from the graph, is approximately x = 0.30. Using the fact that the sum of the two roots of x² − 7x + 2 = 0 is 7, estimate the other root, correct to 2 decimal places.
y = x² − 7x + 2
(a)x ≈ 6.70
(b)x ≈ 6.30
(c)x ≈ 7.30
(d)x ≈ 0.70
21.Solve x² + 7x = 0.
(a)x = 0 or x = 7
(b)x = 0 or x = −14
(c)x = 0 or x = −7
(d)x = −7
22.When a number is added to its square the result is 30. Work out the possible values of the number.
(a)6 and −5
(b)5 and −6
(c)5 and 6
(d)5 only
23.Factorise 6x² − 7x − 3.
(a)(x − 3)(6x + 1)
(b)(3x + 1)(2x − 3)
(c)(6x − 1)(x + 3)
(d)(3x − 1)(2x + 3)
24.A table shows y = x² − 6x + 5 at these points (x, y): (0, 5), (1, 0), (2, −3), (3, −4), (4, −3), (5, 0), (6, 5). Using the symmetry shown, write down the x-coordinate of the turning point.
y = x² − 6x + 5
(a)x = 1
(b)x = 5
(c)x = 6
(d)x = 3

What is on this worksheet?

The sheet holds 24 questions drawn from the MathsUK bank — the content area covered: Algebra (statements A18, A11, A4). It is pitched at GCSE Higher and takes about 45 minutes to work through in full. It is built for independent practice, with full answers at the end for self-marking.

How to use the sheet well

Print it or open it on screen — both work. Printing is A4; the screen view fits phones and tablets.
Do all 24 questions before checking — about 45 minutes is the guide, but there is no time pressure.
Check the answers — press “Show answers” or print the answer page separately.
Redo the questions you got wrong — twice as effective as doing 24 fresh ones.
“New questions” — builds a fresh sheet on the same statements, so you can practise again without repeats.

Why this sheet helps

MathsUK worksheets use questions graded by difficulty and a fair spread of correct-answer positions (the answer is not always (a)) — so the student really has to think about each question rather than guess a pattern. Every question is tagged to a DfE content statement and checked before it enters the bank. The answers come with a step-by-step explanation, not just a value — so a wrong answer becomes a lesson.

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