Pearson Edexcel International Advanced Level Mathematics: Indices, surds and quadratics (YMA01) – 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 follow the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Issue 3, April 2019, Unit P1 topics 1.1 to 1.5: indices, surds, quadratic functions and graphs, the discriminant and completing the square. P1 is a compulsory unit for IAS and IAL Mathematics and Pure Mathematics. Several questions are best done by hand, as the working is the point. Read the study guide first if you need the methods, and use the revision notes to recap. The course hub and printable checklist show where this unit fits, and the free 10-minute diagnostics help you pick what to practise.

Questions

1. Evaluate, without a calculator.

(a) 16^(3/4) [1] (b) (49/9)^(−3/2) [2]

2. Simplify (49x⁸)^(−1/2) × (x^(7/3))³, giving your answer as a single term. [3]

3. Solve each equation.

(a) x^(−3/2) = 1/64 [2] (b) 3x^(5/2) = 96 [2]

4. Simplify, giving exact answers in the form a√b or a + b√c.

(a) √147 − 2√27 + √48 [2] (b) (4 − √6)² [2]

5. Write each expression with a rational denominator.

(a) 15/√12 [2] (b) (2 + √5)/(3 − √5) [3]

6. Show that (√10 + √2)/(√10 − √2) = (3 + √5)/2. [4]

7. Let f(x) = 3 + 2x − x².

(a) Factorise f(x) and hence solve f(x) = 0. [2] (b) Write f(x) in the form p − (x − q)², where p and q are constants. [2] (c) State the coordinates of the turning point of y = f(x) and the equation of its axis of symmetry. Describe the shape of the graph and give the coordinates of the points where it meets the axes. [3]

8. The function f is defined by f(x) = 3x² + 18x + 11.

(a) Express f(x) as a(x + b)² + c, stating the values of a, b and c. [3] (b) Hence solve f(x) = 0, giving your answers as exact values. [3] (c) State the minimum value of f(x) and the value of x at which it occurs. [1]

9. The equation x² − 2kx + (k + 6) = 0, where k is a constant, has equal roots.

(a) Show that k² − k − 6 = 0 and find the possible values of k. [3] (b) For the larger value of k, solve the equation. [1] (c) Find the set of values of k for which the equation has two distinct real roots. [2]

10. The equation x² − (k + 4)x + (2k + 4) = 0, where k is a constant, has discriminant D.

(a) Show that D = k². [3] (b) Hence explain why the equation has real roots for every value of k. State the value of k for which the roots are equal and find the repeated root. [2]

11. The equation x² − 8x − 1 = 0 has a positive root α.

(a) By completing the square, show that α = 4 + √17. [3] (b) Write 1/α in the form p + q√17, where p and q are integers. [3] (c) Hence show that α − 1/α = 8. [2]

Answers

1. (a) 16^(3/4) = (⁴√16)³ = 2³ = 8 [1] [1]

(b) (49/9)^(−3/2) = (9/49)^(3/2) [1] = (3/7)³ = 27/343 [1] [2]

Examiner insight: Work the root before the power. Write down the flip and the root, so that a slip in the final arithmetic still leaves your method visible.

2. (49x⁸)^(−1/2) = 1/(7x⁴) [1]; (x^(7/3))³ = x⁷ [1]; product = x⁷/(7x⁴) = x³/7 [1] [3]

Examiner insight: Apply the power to the number as well as to the letter. A common error is (49x⁸)^(−1/2) = 1/(49x⁴), which gives a wrong answer.

3. (a) x^(−3/2) = 1/64 gives x^(3/2) = 64 [1], so x = 64^(2/3) = 4² = 16 [1] [2]

(b) x^(5/2) = 32 [1], so x = 32^(2/5) = 2² = 4 [1] [2]

Examiner insight: Isolate the power before you raise both sides to the reciprocal exponent. Raising 96 to a power while the 3 is still attached is the usual route to a wrong answer.

4. (a) √147 = 7√3, 2√27 = 6√3, √48 = 4√3 [1]; 7√3 − 6√3 + 4√3 = 5√3 [1] [2]

(b) (4 − √6)² = 16 − 8√6 + 6 [1] = 22 − 8√6 [1] [2]

Examiner insight: Simplify each surd before you collect terms, and expand a squared bracket in full. Writing (4 − √6)² as 16 + 6 drops the middle term and loses credit.

5. (a) √12 = 2√3, so 15/(2√3) [1] = 15√3/6 = 5√3/2 [1] [2]

(b) Multiply by (3 + √5)/(3 + √5) [1]. Numerator: (2 + √5)(3 + √5) = 6 + 2√5 + 3√5 + 5 = 11 + 5√5 [1]. Denominator: 9 − 5 = 4. Result (11 + 5√5)/4 [1] [3]

Examiner insight: Use the conjugate of the denominator only, and state the new denominator as a number. Check that no surd remains on the bottom before moving on.

6. Multiply top and bottom by (√10 + √2) [1]. Numerator: (√10 + √2)² = 10 + 2√20 + 2 = 12 + 4√5 [1]. Denominator: 10 − 2 = 8 [1]. Then (12 + 4√5)/8 = (3 + √5)/2 [1] [4]

Examiner insight: On a “show that” question, write each line of the rationalising and finish on the exact printed result. Simplifying 2√20 to 4√5 is a step worth writing out.

7. (a) f(x) = (3 − x)(1 + x) [1]. So f(x) = 0 gives x = 3 or x = −1 [1] [2]

(b) f(x) = −(x² − 2x) + 3 = −[(x − 1)² − 1] + 3 [1] = 4 − (x − 1)² [1] [2]

(c) The turning point is (1, 4), a maximum [1]. The axis of symmetry is x = 1 [1]. The graph is ∩-shaped (a = −1 < 0) and meets the x-axis at (−1, 0) and (3, 0) and the y-axis at (0, 3) [1] [3]

Examiner insight: Part (b) points you to the turning point in part (c), so use it. Aim for labelled intercepts and a vertex that sits midway between the roots.

8. (a) 3x² + 18x + 11 = 3(x² + 6x) + 11 [1] = 3[(x + 3)² − 9] + 11 [1] = 3(x + 3)² − 16 [1] [3]

(b) 3(x + 3)² = 16, so (x + 3)² = 16/3 [1]. Then x + 3 = ±4/√3 [1]. Rationalising, x = −3 ± 4√3/3 [1] [3]

(c) The minimum value is −16, at x = −3 [1] [1]

Examiner insight: “Hence” means use part (a); solving again with the formula may reach the same answer but can forfeit the method the question asked for. Remember to rationalise 4/√3 as 4√3/3.

9. (a) Equal roots means b² − 4ac = 0, so (−2k)² − 4(1)(k + 6) = 0 [1]. This gives 4k² − 4k − 24 = 0 [1], so k² − k − 6 = 0. Factorising, (k − 3)(k + 2) = 0, so k = 3 or k = −2 [1] [3]

(b) For k = 3 the equation is x² − 6x + 9 = 0, so (x − 3)² = 0 and x = 3 [1] [1]

(c) Two distinct roots need k² − k − 6 > 0. The U-shaped graph of k² − k − 6 crosses the k-axis at −2 and 3 [1], so it is positive when k < −2 or k > 3 [1] [2]

Examiner insight: Write the discriminant with brackets round −2k so that the square is positive. A common slip in part (c) is giving the interval between the roots, which is the “no real roots” case.

10. (a) Here a = 1, b = −(k + 4) and c = 2k + 4 [1]. The discriminant is (k + 4)² − 4(2k + 4) [1] = k² + 8k + 16 − 8k − 16 = k² [1] [3]

(b) Since k² ≥ 0 for every real k, the discriminant is never negative, so the roots are always real [1]. The roots are equal when k = 0, and the equation is then x² − 4x + 4 = 0, so the repeated root is x = 2 [1] [2]

Examiner insight: Say why the discriminant cannot be negative, for example “a square is never negative”, and do not simply assert it. State both the value of k and the root when both are asked for.

11. (a) x² − 8x − 1 = 0 gives (x − 4)² − 16 − 1 = 0 [1]. So (x − 4)² = 17 [1]. Then x = 4 ± √17, and the positive root is α = 4 + √17 since 4 − √17 is negative [1] [3]

(b) 1/α = 1/(4 + √17), multiplied by (4 − √17)/(4 − √17) [1]. The denominator is 16 − 17 = −1 [1]. So 1/α = (4 − √17)/(−1) = −4 + √17, with p = −4 and q = 1 [1] [3]

(c) α − 1/α = (4 + √17) − (−4 + √17) [1] = 8 [1] [2]

Examiner insight: Justify the choice of the positive root rather than quoting both. In part (b), a negative denominator is where most sign slips happen, so write it as its own line.

Where marks are usually lost

• Raising to the fractional power in the wrong order, or forgetting to flip the fraction for a negative power.
• Applying a fractional power to the letter but not to the numerical coefficient in a bracket.
• Dropping the middle term when squaring a bracket that contains a surd.
• Writing a decimal approximation when an exact answer is requested.
• Rationalising with the wrong conjugate or leaving a surd in the denominator.
• Losing a minus sign when the denominator comes out negative.
• Taking out the factor a incorrectly when completing the square, so the constant term is wrong.
• Reading the turning point of (x − 1)² + 4 as (−1, 4).
• Writing the discriminant without brackets round a negative coefficient.
• Giving the “no real roots” interval when the question asks for two distinct roots.

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 Edexcel 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

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019 (first teaching September 2018), published by Pearson Education Limited: Unit P1, topics 1.1 to 1.5.