OxfordAQA International GCSE Mathematics: Solving quadratics: factorising, completing the square (9260) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover reference A20 in section 3.2.3 of the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards). Factorising and graph solutions are Core. Completing the square and the quadratic formula are Extension only, and those questions are labelled “(Extension only)”. The specification says all content can be assessed on the Extension tier question papers, so Core content is also Extension content.

If a method is unfamiliar, start with the study guide. The revision notes give quick recall. The course page is OxfordAQA IGCSE Mathematics and the printable list is the checklist.

Questions

1. Solve x² + 3x − 54 = 0. [3]

2. Solve each equation.

(a) 7x² − 21x = 0 [3] (b) 25x² − 16 = 0 [2]

3. Solve 6x² − 5x − 21 = 0. [4]

4. Solve (x + 2)(x + 5) = 4x + 28. [4]

5. Here is a table of values for y = x² − 4x − 1.

x	−1	0	1	2	3	4	5
y	4	−1	−4	−5	−4	−1	4

(a) Draw the graph of y = x² − 4x − 1 for values of x from −1 to 5. [2] (b) Use your graph to estimate the solutions of x² − 4x − 1 = 0. [2] (c) Write down the equation of a line you could draw on the graph to solve x² − 4x − 3 = 0. Show how you decide, then use your graph to estimate the solutions. [3]

6. (Extension only)

(a) Write x² − 8x + 6 in the form (x + p)² + q. [2] (b) Hence solve x² − 8x + 6 = 0. Give your answers in the form a ± √b. [2]

7. (Extension only) Solve 3x² + 12x − 5 = 0 by completing the square. Give your answers correct to 2 decimal places. [5]

8. (Extension only) Use the quadratic formula to solve 4x² − 9x − 1 = 0. Round your answers to 3 significant figures. [5]

9. A rectangle has length (2x + 1) cm and width (x − 2) cm. Its area is 52 cm².

(a) Show that 2x² − 3x − 54 = 0. [3] (b) Solve 2x² − 3x − 54 = 0. [3] (c) Write down the length and width of the rectangle. [2]

10. (Extension only) A ball is thrown upwards. Its height h metres above the ground after t seconds is h = 2 + 12t − 5t².

(a) Find the time when the ball hits the ground. Give your answer correct to 3 significant figures. [4] (b) Find the times when the ball is 9 metres above the ground. [3]

11. (Extension only) The number x satisfies 3/x + 2/(x + 1) = 2.

(a) Show that 2x² − 3x − 3 = 0. [3] (b) Solve 2x² − 3x − 3 = 0. Round your answers to 3 significant figures. [3]

Answers

1. Two numbers with product −54 and sum 3 are 9 and −6, so (x + 9)(x − 6) = 0 [1]. x = −9 [1]. x = 6 [1]. x = −9 or x = 6 [3] Examiner insight: Write the brackets down before the solutions, so that your method is visible even if a later sign slips.

2. (a) Take out the common factor: 7x(x − 3) = 0 [1]. x = 0 [1]. x = 3 [1]. x = 0 or x = 3 (b) (5x − 4)(5x + 4) = 0 [1]. x = 4/5 or x = −4/5 [1] Examiner insight: Dividing both sides of 7x² = 21x by x gives only x = 3, so the solution x = 0 is lost; factorising keeps both.

3. a × c = −126, and −14 and 9 have product −126 and sum −5. 6x² − 14x + 9x − 21 = 0 [1]. 2x(3x − 7) + 3(3x − 7) = 0 [1]. (3x − 7)(2x + 3) = 0 [1]. x = 7/3 or x = −3/2 [1] [4] Examiner insight: Expanding your brackets to check they give 6x² − 5x − 21 takes seconds and catches sign errors before they reach the solutions.

4. Expand: x² + 7x + 10 = 4x + 28 [1]. Rearrange: x² + 3x − 18 = 0 [1]. (x + 6)(x − 3) = 0 [1]. x = −6 or x = 3 [1] [4] Examiner insight: Setting x + 2 and x + 5 equal to 4x or 28 does not work, because the right-hand side is not 0; this approach gives wrong solutions.

5. (a) All seven points plotted correctly [1]. A smooth curve through them, with a rounded minimum [1]. [2] (b) The curve crosses y = 0 between x = −1 and x = 0, and between x = 4 and x = 5 [1]. x ≈ −0.2 and x ≈ 4.2 (answers within about 0.1 of these are reasonable) [1]. [2] (c) x² − 4x − 3 = 0 becomes x² − 4x − 1 = 2 [1], so the line is y = 2 [1]. Reading where it meets the curve gives x ≈ −0.6 and x ≈ 4.6 [1]. [3] Examiner insight: Solutions are x-values, so mark them clearly on the x-axis or list them as x = …; quoting the y-values of the points is a common way to lose the final mark.

6. (a) (x − 4)² [1] − 10, because −16 + 6 = −10 [1]. (x − 4)² − 10 [2] (b) (x − 4)² = 10, so x − 4 = ±√10 [1]. x = 4 ± √10 [1] [2] Examiner insight: The form a ± √b asks for an exact answer, so giving only decimals does not match what was requested.

7. 3x² + 12x − 5 = 3(x² + 4x) − 5 [1] = 3[(x + 2)² − 4] − 5 = 3(x + 2)² − 17 [1]. So 3(x + 2)² = 17 and (x + 2)² = 17/3 [1]. x = −2 ± √(17/3) [1]. x = 0.38 or x = −4.38 [1] [5] Examiner insight: When the question names the method, an answer found by the formula alone does not show the method asked for, so show the completed-square form.

8. a = 4, b = −9, c = −1 [1]. b² − 4ac = 81 + 16 = 97 [1]. x = (9 ± √97)/8 [1]. x = 2.36 [1] or x = −0.106 [1] [5] Examiner insight: Writing the substitution with brackets, as (−9)², prevents the common slip of getting −81; a wrong discriminant is easier to spot when it is shown on its own line.

9. (a) (2x + 1)(x − 2) = 2x² − 4x + x − 2 [1] = 2x² − 3x − 2 [1]. Setting this equal to 52 gives 2x² − 3x − 54 = 0 [1]. [3] (b) 2x² − 12x + 9x − 54 = 0 [1]. (2x + 9)(x − 6) = 0 [1]. x = 6 or x = −4.5 [1] [3] (c) A width of x − 2 must be positive, so x = 6 [1]. Length 13 cm and width 4 cm [1] (check: 13 × 4 = 52). [2] Examiner insight: A “show that” needs every line, including the expansion; and a context question usually expects the invalid solution to be rejected with a reason.

10. (a) At the ground h = 0, so 5t² − 12t − 2 = 0 [1]. t = (12 ± √(144 + 40))/10 = (12 ± √184)/10 [1]. t = 2.56 or t = −0.156 [1]. Time cannot be negative: t = 2.56 seconds [1] [4] (b) 2 + 12t − 5t² = 9 gives 5t² − 12t + 7 = 0 [1]. (5t − 7)(t − 1) = 0 [1]. t = 1 second and t = 1.4 seconds [1] [3] Examiner insight: Two answers in part (b) make sense, since the ball passes 9 metres on the way up and on the way down; giving one value suggests the second root was overlooked.

11. (a) Multiply every term by x(x + 1): 3(x + 1) + 2x = 2x(x + 1) [1]. Expand: 3x + 3 + 2x = 2x² + 2x [1]. Rearrange: 2x² − 3x − 3 = 0 [1]. [3] (b) a = 2, b = −3, c = −3, so b² − 4ac = 9 + 24 = 33 [1]. x = (3 ± √33)/4 [1]. x = 2.19 or x = −0.686 [1] [3] Examiner insight: Both solutions are valid here because neither makes a denominator zero; it is good practice to check this rather than assume.

Where marks are usually lost

• Solving before the equation is in the form “= 0”, for example setting each bracket equal to a non-zero number.
• Dividing by x and losing the solution x = 0.
• Wrong signs when splitting the middle term of a trinomial with a ≠ 1.
• Stopping at the factorised form and not stating the values of x.
• Reading y-coordinates instead of x-coordinates from a graph.
• Choosing the wrong horizontal line when the equation has been rearranged.
• Forgetting to take out the coefficient of x² before halving when completing the square.
• Writing −b² for (−b)² in the formula when b is negative.
• Rounding early, or giving a decimal when a surd was requested.
• Leaving a negative length or time in a context question without rejecting it.

Next steps

• Review the revision notes and the study guide.
• Return to the OxfordAQA IGCSE Mathematics course page and tick off topics on the checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA. Section 3.2.3, reference A20.