AQA A-level Mathematics: Rational Indices and Surds (7357) – 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.

This set covers content references B1 (laws of indices for all rational exponents) and B2 (surds, including rationalising the denominator) from section 3.3 B: Algebra and functions of the AQA A-level Mathematics (7357) specification (version 1.3, dated 31 January 2018), which applies to A-level exams from June 2018. 7357 is a linear A-level with no AS/A split. Section B is Paper 1 content, and Papers 2 and 3 can assess any Paper 1 content. Work every question by hand and show each step, as you would in an exam.

Read the study guide first and use the revision notes to recall the methods. Course links: AQA A-level Mathematics hub and printable checklist.

Questions

1. Evaluate, giving each answer as an integer or a fraction.

(a) 32^(3/5) [1] (b) 81^(−3/4) [1] (c) (1/4)^(−3/2) [1]

2. Simplify each expression, where x > 0.

(a) (27x⁶)^(2/3) × x⁻⁴ [2] (b) √x × x^(5/2) ÷ x^(3/2) [2]

3. Write each in the form k√n, where n is as small as possible.

(a) √75 − √12 [2] (b) √18 × √10 [2]

4. Rationalise the denominator and simplify.

(a) 14/√7 [1] (b) 8/(5 − √17) [2] (c) 3/(2√3 − 3) [2]

5. Solve each equation.

(a) 5^(2x−1) = 125^(x−2) [3] (b) 2ˣ × 4^(1−x) = ½ × 8ˣ [3]

6. Given that x > 0, solve each equation.

(a) x^(2/3) = 16 [2] (b) x^(−3/2) = 1/27 [2]

7. Show that (3 + √5)²/(3 − √5) = 18 + 8√5. [5]

8. The equation 25ˣ − 6 × 5ˣ + 5 = 0 is to be solved.

(a) Show that, with y = 5ˣ, the equation becomes y² − 6y + 5 = 0. [2] (b) Hence find the values of x. [4]

9. A right-angled triangle has a base of (2 + √2) cm and an area of (7 + 3√2) cm². The base is one of the two sides next to the right angle.

(a) Find the height of the triangle in the form a + b√2, where a and b are integers. [4] (b) Find the exact value of the square of the length of the hypotenuse, in the form c + d√2. [2]

10. Write (5 + √3)/(2 − √3) in the form p + q√3, where p and q are integers. [4]

11. Show that 1/(1 + √2) + 1/(√2 + √3) + 1/(√3 + 2) = 1. [5]

12. Solve 2x − 5√x + 2 = 0. [5]

Answers

1. (a) 32^(3/5) = (⁵√32)³ = 2³ = 8 [1] (b) 81^(−3/4) = 1/(⁴√81)³ = 1/3³ = 1/27 [1] (c) (1/4)^(−3/2) = 4^(3/2) = (√4)³ = 2³ = 8 [1] Examiner insight: Take the root before the power so the numbers stay small. Writing the root step, as above, shows the method even if you slip on the final arithmetic.

2. (a) (27x⁶)^(2/3) = 27^(2/3) × x⁴ = 9x⁴ [1]. Then 9x⁴ × x⁻⁴ = 9x⁰ = 9 [1] (b) Add and subtract the indices: 1/2 + 5/2 − 3/2 = 3/2 [1]. The result is x^(3/2) [1] Examiner insight: Apply the fractional power to the number as well as to the letter. Leaving 27 untouched is the most common way to lose the first mark.

3. (a) √75 = 5√3 and √12 = 2√3 [1]. So √75 − √12 = 3√3 [1] (b) √18 × √10 = √180 [1] = √36 × √5 = 6√5 [1] Examiner insight: Simplify each surd separately before you combine. Always take out the largest square factor, otherwise the form k√n is not complete (for example 3√20 would need a further step).

4. (a) 14/√7 = 14√7/7 = 2√7 [1] (b) Multiply top and bottom by (5 + √17). The denominator is 25 − 17 = 8 [1]. So 8(5 + √17)/8 = 5 + √17 [1] (c) Multiply top and bottom by (2√3 + 3). The denominator is (2√3)² − 3² = 12 − 9 = 3 [1]. So 3(2√3 + 3)/3 = 2√3 + 3 [1] Examiner insight: State the conjugate you are using, then use the difference of two squares on the denominator. A numerator multiplied by a different quantity from the denominator changes the value, so check both lines match.

5. (a) 125 = 5³, so 125^(x−2) = 5^(3x−6) [1]. Equate the indices: 2x − 1 = 3x − 6 [1]. x = 5 [1] (b) Left side: 2ˣ × 2^(2−2x) = 2^(2−x) [1]. Right side: ½ × 2^(3x) = 2^(3x−1) [1]. Then 2 − x = 3x − 1, so 4x = 3 and x = 3/4 [1] Examiner insight: Reduce both sides to a single power of the same base before equating indices. Do not mix methods halfway through, and check the answer by substituting back.

6. (a) Raise both sides to the power 3/2: x = 16^(3/2) [1] = (√16)³ = 64 [1] (b) Rearrange: x^(3/2) = 27 [1]. Raise to the power 2/3: x = 27^(2/3) = 3² = 9 [1] Examiner insight: Use the reciprocal of the index to undo the power on x. Mentioning the condition x > 0 is sensible, because it removes any doubt about negative roots.

7. Expand the numerator: (3 + √5)² = 9 + 6√5 + 5 = 14 + 6√5 [1]. Multiply top and bottom by (3 + √5) [1]. Numerator: (14 + 6√5)(3 + √5) = 42 + 14√5 + 18√5 + 30 = 72 + 32√5 [1]. Denominator: (3 − √5)(3 + √5) = 9 − 5 = 4 [1]. So the fraction is (72 + 32√5)/4 = 18 + 8√5 [1] Examiner insight: In a “show that” question, every step must be visible and you must end on the printed result. Do not work backwards from 18 + 8√5.

8. (a) 25ˣ = (5²)ˣ = (5ˣ)² = y² [1]. So the equation becomes y² − 6y + 5 = 0 [1] (b) Factorise: (y − 1)(y − 5) = 0 [1]. So y = 1 or y = 5 [1]. Then 5ˣ = 1 gives x = 0 [1] and 5ˣ = 5 gives x = 1 [1] Examiner insight: The question asks for x, not y. Convert back, and note that both values of y are positive, so both are acceptable.

9. (a) Area = ½ × base × height, so height = 2(7 + 3√2)/(2 + √2) = (14 + 6√2)/(2 + √2) [1]. Multiply top and bottom by (2 − √2) [1]. Numerator: (14 + 6√2)(2 − √2) = 28 − 14√2 + 12√2 − 12 = 16 − 2√2. Denominator: 4 − 2 = 2 [1]. Height = (8 − √2) cm [1] (b) (2 + √2)² = 6 + 4√2 and (8 − √2)² = 66 − 16√2 [1]. By Pythagoras the square of the hypotenuse is 72 − 12√2 [1] Examiner insight: Start from the area formula and write the division as a single fraction. In part (b), the middle terms of both squares are easy to drop, so expand each bracket in full.

10. Multiply top and bottom by the conjugate (2 + √3) [1]. Numerator: (5 + √3)(2 + √3) = 10 + 5√3 + 2√3 + 3 = 13 + 7√3 [1]. Denominator: (2 − √3)(2 + √3) = 4 − 3 = 1 [1]. So the value is 13 + 7√3, with p = 13 and q = 7 [1] Examiner insight: A denominator of 1 is a good sign and a useful check. State p and q explicitly when the question names them.

11. First term: 1/(1 + √2) × (√2 − 1)/(√2 − 1) = (√2 − 1)/(2 − 1) = √2 − 1 [1] Second term: (√3 − √2)/(3 − 2) = √3 − √2 [1] Third term: (2 − √3)/(4 − 3) = 2 − √3 [1] Add the three results: (√2 − 1) + (√3 − √2) + (2 − √3) [1] The surds cancel in pairs, leaving 2 − 1 = 1 [1] Examiner insight: Write all three rationalised terms before adding. The cancelling pattern then shows the result clearly. Ending on 1 is part of the proof.

12. Let u = √x, so x = u² [1]. The equation becomes 2u² − 5u + 2 = 0 [1]. Factorise: (2u − 1)(u − 2) = 0 [1]. So u = 1/2 or u = 2, and both are positive, so both are valid [1]. Square: x = 1/4 or x = 4 [1] Examiner insight: This is a quadratic in √x. Giving u as the final answer, or forgetting to square, is the usual error, so finish by returning to x.

Where marks are usually lost

• Adding indices when the bases are different, for example 2³ × 3² treated as 6⁵.
• Applying a fractional index to the letter but not to the number in the bracket.
• Treating a negative index as a negative number instead of a reciprocal.
• Not reducing both sides of an index equation to the same base before equating the indices.
• Using the wrong sign in the conjugate, or multiplying only the denominator.
• Dropping the middle term when expanding a bracket that contains a surd, such as (3 + √5)².
• Rounding a surd to a decimal when the question wants an exact form.
• Solving the quadratic in y = aˣ or in u = √x and not converting back to x.
• Missing steps in a “show that” question, or working backwards from the printed answer.

Next steps

• Recap the methods with the revision notes.
• Re-read the worked examples in the study guide.
• Tick off the topic on the printable checklist and browse the AQA A-level Mathematics hub.
• Find your weak spots with the free 10-minute diagnostics.
• Want help from a tutor? Book a free trial class.

Official syllabus

AQA, A-level Mathematics 7357, specification (version 1.3, dated 31 January 2018), which applies to A-level exams from June 2018: section 3.3 B: Algebra and functions, content references B1 and B2.