OxfordAQA International AS and A-level Mathematics: Surds, indices, quadratics, simultaneous equations (9660) – 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 specification content P1.1 (Algebra) in the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2: surds, indices, quadratics, simultaneous equations and inequalities. The specification states this content is part of International AS Unit P1 (Pure Maths, section 3.1). Learn the methods first in the study guide or the revision notes.

Related links: OxfordAQA Mathematics hub, printable checklist, Pure Maths 1 practice questions and free 10-minute diagnostics.

Questions

1. Simplify √98 − √32 + √18, giving your answer in the form k√2. [3]

2. Rationalise the denominator of 20/(6 − √11), simplifying your answer fully. [3]

3. Show that (2 + √7)/(4 − √7) = (5 + 2√7)/3. [4]

4.

(a) Evaluate (125/27)^(−2/3). [2] (b) Solve 4^(x + 1) = 8^(x − 2). [2]

5. Solve 6x² + 11x − 35 = 0. [3]

6. f(x) = 2x² − 8x + 11.

(a) Write f(x) in the form a(x + p)² + q. [3] (b) State the vertex of the graph of y = f(x) and the equation of its line of symmetry. [2] (c) Explain why the graph of y = f(x) does not cross the x-axis. [1]

7. Solve 3x² − 8x + 2 = 0, giving your answers in the form (a ± √b)/c. [3]

8. The equation x² + kx + (k + 3) = 0 has real coefficients and k is a constant.

(a) Find the values of k for which the equation has equal roots. [3] (b) Determine the range of k that gives the equation two distinct real roots. [3]

9. Solve the simultaneous equations 2x + y = 3 and x² + y² = 5. [6]

10.

(a) Solve 3(2 − x) ≥ 4x − 15. [2] (b) Solve 4x² + x − 18 > 0. [3] (c) Hence find all values of x that satisfy both 3(2 − x) ≥ 4x − 15 and 4x² + x − 18 > 0. [2]

11. The line y = 2x + k and the curve y = x² − 4x + 5 are drawn on the same axes.

(a) Show that any intersection point has an x-coordinate satisfying x² − 6x + 5 − k = 0. [2] (b) The line is a tangent to the curve. Find the value of k. [3] (c) For k = −1, find the exact coordinates of the points where the line meets the curve. [3]

12. A farmer uses a straight wall as one side of a rectangular enclosure. The other three sides use 40 m of fencing. The sides perpendicular to the wall are each x metres long and the enclosed area is A m².

(a) Show that A = 40x − 2x². [1] (b) Write A in the form p − 2(x − q)². Hence state the largest area available and the width that produces it. [3] (c) The farmer needs an area of at least 150 m². Find the possible values of x. [3] (d) Find, in the form m ± n√10, the values of x that give an area of exactly 120 m². [3]

Answers

1. √98 = 7√2 [1]; √32 = 4√2 and √18 = 3√2 [1]; 7√2 − 4√2 + 3√2 = 6√2 [1] [3] Examiner insight: Show each simplification on its own line. Working that is laid out clearly makes a slip easy to spot and fix, whereas a bare total gives you nothing to check.

2. Multiply top and bottom by (6 + √11) [1]; denominator 36 − 11 = 25 [1]; 20(6 + √11)/25 = (24 + 4√11)/5 [1] [3] Examiner insight: “Simplifying fully” means cancelling the common factor of 5 from 20 and 25. Stopping at 20(6 + √11)/25 leaves the answer unfinished.

3. Multiply top and bottom by (4 + √7) [1]; numerator (2 + √7)(4 + √7) = 8 + 2√7 + 4√7 + 7 = 15 + 6√7 [1]; denominator 16 − 7 = 9 [1]; (15 + 6√7)/9 = (5 + 2√7)/3, as required [1] [4] Examiner insight: On a “show that” question, every stage must be visible and you must reach the stated result. Working backwards from the printed result is not a safe route to full credit.

4. (a) (125/27)^(−2/3) = (27/125)^(2/3) [1]; (3/5)² = 9/25 [1] (b) 2^(2x + 2) = 2^(3x − 6) [1]; 2x + 2 = 3x − 6, so x = 8 [1] [4] Examiner insight: Convert to a common base before comparing exponents. Taking the reciprocal first, then the root, then the power keeps part (a) free of large numbers.

5. Need two numbers with product −210 and sum 11: 21 and −10 [1]; (3x − 5)(2x + 7) = 0 [1]; x = 5/3 or x = −7/2 [1] [3] Examiner insight: Quote both roots. A correct factorisation with only one root given leaves the question unfinished. The formula is a fine alternative if you are careful with the signs.

6. (a) 2(x² − 4x) + 11 [1]; 2[(x − 2)² − 4] + 11 [1]; 2(x − 2)² + 3 [1] (b) Vertex (2, 3) [1]; line of symmetry x = 2 [1] (c) The minimum value of y is 3, which is above the x-axis (and a > 0 so the parabola opens upward) [1] [6] Examiner insight: Take the 2 out of both terms before completing the square. For part (c), refer to the minimum value; the discriminant, 64 − 88 = −24, is another valid way to justify it.

7. b² − 4ac = 64 − 24 = 40 [1]; x = (8 ± √40)/(2 × 3) [1]; √40 = 2√10, so x = (4 ± √10)/3 [1] [3] Examiner insight: Write the formula with values substituted, in brackets where a number is negative. This makes slips easier to find. Simplify √40 before cancelling, and cancel the common factor of 2 from every term.

8. (a) b² − 4ac = k² − 4(k + 3) = k² − 4k − 12 [1]; equal roots means (k − 6)(k + 2) = 0 [1]; k = 6 or k = −2 [1] (b) Distinct real roots needs k² − 4k − 12 > 0 [1]; (k − 6)(k + 2) > 0 [1]; k < −2 or k > 6 [1] [6] Examiner insight: A set of values is an inequality, not an equation. Do not write a single chain such as −2 > k > 6, and keep the two regions joined by “or”.

9. y = 3 − 2x [1]; x² + (3 − 2x)² = 5 [1]; 5x² − 12x + 4 = 0 [1]; (5x − 2)(x − 2) = 0 [1]; x = 2/5 or x = 2 [1]; y = 3 − 2x gives y = 11/5 and y = −1, so (2/5, 11/5) and (2, −1) [1] [6] Examiner insight: Substitute back into the linear equation, which gives one y for each x. Using the quadratic to find y can produce wrong pairs, and the answer should be presented as pairs.

10. (a) 6 − 3x ≥ 4x − 15 [1]; 21 ≥ 7x, so x ≤ 3 [1] (b) (x − 2)(4x + 9) > 0 [1]; critical values x = 2 and x = −9/4 [1]; upward parabola, so x < −9/4 or x > 2 [1] (c) x < −9/4 [1]; 2 < x ≤ 3 [1] [7] Examiner insight: Part (b) is helped by a quick sketch or sign test, to decide between “outside” and “between”. For part (c), draw both solution sets on one number line, then check whether each end point is included.

11. (a) x² − 4x + 5 = 2x + k [1]; x² − 6x + 5 − k = 0, as required [1] (b) b² − 4ac = 36 − 4(5 − k) = 16 + 4k [1]; tangent means one repeated root, so 16 + 4k = 0 [1]; k = −4 [1] (c) x² − 6x + 6 = 0 [1]; x = (6 ± √12)/2 = 3 ± √3 [1]; y = 2x − 1 gives (3 + √3, 5 + 2√3) and (3 − √3, 5 − 2√3) [1] [8] Examiner insight: In part (a) the final equation must appear explicitly. In part (c), “exact” rules out decimals, so keep the surds, and find y from the line.

12. (a) Length parallel to the wall is 40 − 2x, so A = x(40 − 2x) = 40x − 2x² [1] (b) −2(x² − 20x) [1]; = 200 − 2(x − 10)², so p = 200 and q = 10 [1]; greatest area 200 m² when x = 10 [1] (c) 40x − 2x² ≥ 150, so x² − 20x + 75 ≤ 0 [1]; (x − 5)(x − 15) ≤ 0 [1]; 5 ≤ x ≤ 15 [1] (d) 2x² − 40x + 120 = 0, so x² − 20x + 60 = 0 [1]; x = (20 ± √160)/2 with √160 = 4√10 [1]; x = 10 ± 2√10, and both lie between 0 and 20 so both are valid [1] [10] Examiner insight: In context questions, state the limits on x (here 0 < x < 20) and check each answer against them. The completed-square form in part (b) also tells you the maximum without any further working.

Where marks are usually lost

• Leaving a common factor uncancelled, such as 20(6 + √11)/25, after rationalising.
• Sign errors in the conjugate, or squaring a bracket with surds and forgetting the middle term.
• Working out the power before the root, then dealing with huge numbers.
• Taking x² and x terms into a bracket without removing the coefficient a first.
• Giving only one root of a quadratic, or answering a “set of values” question with an equation.
• Mixing up “outside the roots” and “between the roots” for quadratic inequalities.
• Finding the second variable from the quadratic instead of the linear equation, producing false pairs.
• Giving decimals when the question says “exact”, “in surd form” or “in the form m ± n√10”.
• Ignoring the context’s limits on a variable, and then accepting an impossible value.

Next steps

• Review the revision notes and the study guide.
• Go to the OxfordAQA Mathematics hub and the printable checklist.
• Try all the free 10-minute diagnostics to see which topics need work.
• Book a free trial class for support with anything you could not finish.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA. Content P1.1 (Algebra), International AS Unit P1 (Pure Maths).