Edexcel International GCSE Maths Indices and Surds Past Paper Questions and Solutions (4MA1 Higher)

Part (a)

1. √2 = 2^½ and 16 = 2⁴
2. 2^½ × 2⁴ = 2^4.5

Answer: x = 4.5

Part (b)

1. (11⁻⁶)⁵ = 11⁻³⁰ (multiply the indices)
2. 11⁻³⁰ ÷ 11⁴ = 11^{−30 − 4} = 11⁻³⁴

Answer: n = −34

Part (a)

1. A power of ½ is a square root: √16 = 4, (e¹⁰)^½ = e⁵, (f⁶)^½ = f³

Answer: 4e⁵f³

Part (b)

1. Common denominator 12: ^{3(2x + 1) + 4(x − 2)}⁄₁₂
2. Numerator: 6x + 3 + 4x − 8 = 10x − 5

Answer: ^{10x − 5}⁄₁₂

Part (c)

1. Write both sides as powers of 2: 4^k+3 = (2²)^k+3 = 2^{2k + 6} and 16 × 2^k = 2⁴ × 2^k = 2^{k + 4}
2. The powers must be equal: 2k + 6 = k + 4

Answer: k = −2

Part (a)

1. Add the indices: ½ + ²⁄₅ = ⁵⁄₁₀ + ⁴⁄₁₀ = ⁹⁄₁₀

Answer: m = ⁹⁄₁₀

Part (b)

1. Subtract the indices: −10 − (−4) = −6

Answer: n = −6

1. Write 32 as a power of 2: 32 = 2⁵
2. So the powers are equal: −4x = 5

Answer: x = −⁵⁄₄ = −1.25

Part (a)

1. Let x = 0.5727272… Then 1000x = 572.727272… and 10x = 5.727272…
2. Subtract: 990x = 567, so x = ⁵⁶⁷⁄₉₉₀
3. Divide top and bottom by 9: ⁶³⁄₁₁₀

Answer: 990x = 567, so x = ⁵⁶⁷⁄₉₉₀ = ⁶³⁄₁₁₀

Part (b)

1. Multiply top and bottom by 2 + √y: ^{3(2 + √y)}⁄_{(2 − √y)(2 + √y)}
2. The denominator is 4 − y.

Answer: ^{6 + 3√y}⁄_{4 − y}

Part (a)

1. Let x = 4.575757… The block “57” has 2 digits, so 100x = 457.575757…
2. Subtract: 100x − x = 457.5757… − 4.5757…, so 99x = 453 and x = ⁴⁵³⁄₉₉ = ¹⁵¹⁄₃₃
3. 151 ÷ 33 = 4 remainder 19

Answer: x = ¹⁵¹⁄₃₃ = 4¹⁹⁄₃₃

Part (b)

1. Multiply top and bottom by 6 + 3√2: denominator = 36 − (3√2)² = 36 − 18 = 18
2. Numerator: 2(6 + 3√2) = 12 + 6√2
3. ^{12 + 6√2}⁄₁₈ = ^{2 + √2}⁄₃ (dividing top and bottom by 6)

Answer: ^{2 + √2}⁄₃, so a = 2 and b = 3

Part (a)

1. Write everything as a power of 2. √2 = 2^½
2. 8³ = (2³)³ = 2⁹
3. 16^3⁄₂ = (√16)³ = 4³ = 64 = 2⁶
4. So ⁸³⁄_16^3⁄2 = 2^{9 − 6} = 2³
5. 2^½ ÷ 2³ = 2^{½ − 3} = 2^−2.5

Answer: n = −2.5

Part (b)

1. 4% = 0.04, so work out 0.04 × 4.5 × 10¹⁵⁷
2. 0.04 × 4.5 = 0.18, giving 0.18 × 10¹⁵⁷
3. 0.18 = 1.8 × 10⁻¹, so 0.18 × 10¹⁵⁷ = 1.8 × 10¹⁵⁶

Answer: 1.8 × 10¹⁵⁶

Part (a)

1. (√3)² = 3, so (√3)⁵ = (√3)² × (√3)² × √3 = 3 × 3 × √3 = 9√3

Answer: k = 9

Part (b)

1. Rationalise: multiply top and bottom by 3 + √2.
2. Denominator: (3 − √2)(3 + √2) = 9 − 2 = 7
3. Numerator: 21(3 + √2) = 63 + 21√2
4. ^{63 + 21√2}⁄₇ = 9 + 3√2
5. 3√2 = √9 × 2 = √18

Answer: 9 + √18

1. Rationalise the first fraction: multiply top and bottom by √2 + 1. The denominator becomes (√2)² − 1 = 2 − 1 = 1.
2. So ¹²⁄_{√2 − 1} = 12(√2 + 1) = 12√2 + 12
3. (√2)⁵ = (√2)² × (√2)² × √2 = 2 × 2 × √2 = 4√2
4. 12√2 + 12 − 4√2 = 8√2 + 12
5. 2√32 = 2 × √16 × 2 = 2 × 4√2 = 8√2, so 8√2 + 12 = 2√32 + 12 ✓

Answer: 8√2 + 12 = 2√32 + 12

Part (a)

1. Expand: (6 + 2√12)² = 36 + 2 × 6 × 2√12 + (2√12)² = 36 + 24√12 + 4 × 12
2. = 84 + 24√12
3. √12 = √4 × 3 = 2√3, so 24√12 = 48√3
4. 84 + 48√3 = 12(7 + 4√3)

Answer: 84 + 48√3 = 12(7 + 4√3)

Part (b)

1. The negative power flips the fraction: (^t15⁄_27a¹²)^2⁄₃
2. The power ²⁄₃ means cube root, then square. ∛27 = 3, so 3² = 9. (a¹²)^2⁄₃ = a⁸ and (t¹⁵)^2⁄₃ = t¹⁰

Answer: ^t10⁄_9a⁸

1. To rationalise the denominator 3 − √5, multiply top and bottom by 3 + √5 (the same terms with the opposite sign).
2. Denominator: (3 − √5)(3 + √5) = 9 − 5 = 4 (the √5 terms cancel)
3. Numerator: (1 + √5)(3 + √5) = 3 + √5 + 3√5 + 5 = 8 + 4√5
4. So the fraction is ^{8 + 4√5}⁄₄ = 2 + √5

Answer: 2 + √5 (so a = 2, b = 5)

Part (a)

1. A fourth root is the power ¹⁄₄, so ∜[k¹²] = k^{12 ÷ 4} = k³
2. (k³)⁵ = k^{3 × 5} = k¹⁵

Answer: n = 15

Part (b)

1. Rationalise the denominator: multiply top and bottom by 2 + √3.
2. Denominator: (2 − √3)(2 + √3) = 4 − 3 = 1
3. Numerator: 7(2 + √3) = 14 + 7√3
4. Write 7√3 as a single root: 7√3 = √49 × 3 = √147

Answer: √147 + 14

1. Expand: (8 − √x)(5 + √x) = 40 + 8√x − 5√x − x = (40 − x) + 3√x
2. Compare with 21 + y√x. Because x is prime, √x is not a whole number, so the whole-number parts must match and the √x parts must match.
3. 40 − x = 21, so x = 19. And y = 3.

Answer: x = 19, y = 3

1. Expand: √3x − 2 × 3 = x + 2√3, so √3x − 6 = x + 2√3
2. Collect the x terms: √3x − x = 6 + 2√3, so x(√3 − 1) = 6 + 2√3
3. x = ^{6 + 2√3}⁄_{√3 − 1}. Rationalise: multiply top and bottom by (√3 + 1)
4. Top: (6 + 2√3)(√3 + 1) = 6√3 + 6 + 6 + 2√3 = 12 + 8√3. Bottom: 3 − 1 = 2
5. x = ^{12 + 8√3}⁄₂

Answer: x = 6 + 4√3

1. Multiply top and bottom by 3 + √7: ^{6(3 + √7)}⁄_{(3 − √7)(3 + √7)}
2. Denominator: 9 − 7 = 2
3. ^{6(3 + √7)}⁄₂ = 3(3 + √7)

Answer: 9 + 3√7

Part (a)

1. ¹⁄_x = x⁻¹, and a cube root is the power ¹⁄₃: (x⁻¹)^1⁄₃ = x^−1⁄₃
2. Raise to the power 4: x^−4⁄₃

Answer: m = −⁴⁄₃

Part (b)

1. 3(x² + 4x) + 19 = 3[(x + 2)² − 4] + 19 = 3(x + 2)² − 12 + 19

Answer: 3(x + 2)² + 7

1. Write each number in prime factors: 18 = 2 × 3², so 18⁴ⁿ = 2⁴ⁿ × 3⁸ⁿ. And 12² = (2² × 3)² = 2⁴ × 3².
2. Powers of 2: 4n + 3n² − 18n − 4 = 3n² − 14n − 4
3. Powers of 3: 8n + 2 − 8n − 2 = 0, so the 3s cancel completely.
4. M = 2^{3n2 − 14n − 4} = 2¹, so 3n² − 14n − 4 = 1
5. 3n² − 14n − 5 = 0, which factorises as (3n + 1)(n − 5) = 0

Answer: n = 5 or n = −¹⁄₃

1. Simplify √8 = √4 × 2 = 2√2, so the numerator is 3 + 2√2.
2. Expand the denominator: (√2 − 1)² = 2 − 2√2 + 1 = 3 − 2√2
3. Rationalise: multiply top and bottom by 3 + 2√2. The denominator becomes 9 − (2√2)² = 9 − 8 = 1.
4. Numerator: (3 + 2√2)² = 9 + 12√2 + 8 = 17 + 12√2
5. 12√2 = √144 × 2 = √288

Answer: 17 + √288

1. Write everything as a power of 2: 16^x = 2^4x and 8 = 2³
2. 2ⁿ = 2^{x2 + 4x + 3}, so n = x² + 4x + 3
3. Complete the square: x² + 4x + 3 = (x + 2)² − 1, so (x + 2)² = n + 1
4. x + 2 = ±√n + 1. Since x > 0, take the positive root: x = −2 + √n + 1
5. For x > 0 we need √n + 1 > 2, so n + 1 > 4 and n > 3

Answer: x = √n + 1 − 2, with n > 3