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1. (x + 3)² = x² + 6x + 9, which is 4 more than x² + 6x + 5 (M1)
2. So x² + 6x + 5 = (x + 3)² − 4 (A1)

Answer: (x + 3)² − 4

1. Half of −8 is −4, so start with (x − 4)² = x² − 8x + 16 (M1)
2. Subtract 16 and then add the original 3: x² − 8x + 3 = (x − 4)² − 16 + 3 (M1)
3. = (x − 4)² − 13 (A1)

Answer: (x − 4)² − 13

1. (x + 2)² = x² + 4x + 4, so x² + 4x − 1 = (x + 2)² − 5 (M1)
2. Completed square form: (x + 2)² − 5 (A1)
3. Set (x + 2)² − 5 = 0, so (x + 2)² = 5 and x + 2 = ±√5 (M1)
4. x = −2 ± √5 (A1)

Answer: (x + 2)² − 5, so x = −2 ± √5

1. Complete the square: y = (x − 5)² − 25 + 27 (M1)
2. y = (x − 5)² + 2 (A1)
3. The bracket is zero when x = 5, so the minimum point is (5, 2) (A1)
4. The minimum value of y is 2 (A1)

Answer: Minimum point (5, 2), minimum value y = 2

1. Take out the 2 from the first two terms: 2(x² + 4x) + 5 (M1)
2. Complete the square inside the bracket: 2((x + 2)² − 4) + 5 (M1)
3. Multiply out: 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3 (A1)

Answer: 2(x + 2)² − 3

1. From the completed form: 2(x + 2)² − 3 = 0 (M1)
2. So (x + 2)² = 3/2 and x + 2 = ±√(3/2) (M1)
3. √(3/2) = √6 ÷ 2, so x = −2 + (√6 ÷ 2) (A1)
4. The second root is x = −2 − (√6 ÷ 2) (A1)

Answer: x = −2 ± (√6 ÷ 2)

1. Take out the 2: 2(x² − 5x) + 7 = 2((x − 5/2)² − 25/4) + 7 (M1)
2. Multiply out: 2(x − 5/2)² − 25/2 + 7 = 2(x − 5/2)² − 11/2 (M1)
3. Set equal to zero: (x − 5/2)² = 11/4, so x = 5/2 ± √11 ÷ 2 (A1)
4. Write over one denominator: x = (5 ± √11) ÷ 2 (A1)

Answer: x = (5 ± √11) ÷ 2

1. (a) Subtract the powers: 6⁸ ÷ 6⁵ = 6^{8 − 5} = 6³ (M1)
2. (b) Multiply the powers: (3²)⁴ = 3^{2 × 4} = 3⁸ (M1)
3. 6³ = 216 and 3⁸ = 6561 (A1)

Answer: (a) 216 (b) 6561

1. (a) Any non-zero number to the power 0 is 1, so 7⁰ = 1 (A1)
2. (b) A negative power gives a reciprocal: 5⁻² = 1/5² (M1)
3. (b) 1/5² = 1/25 (A1)
4. (c) Invert the fraction: (2/5)⁻¹ = 5/2 (A1)

Answer: (a) 1 (b) 1/25 (c) 5/2

1. (a) A power of 1/2 is a square root: 49^1/2 = √49 = 7 (B1)
2. (b) A power of 1/3 is a cube root: 27^1/3 = 3 (B1)
3. (c) 16^3/4 = (16^1/4)³ = 2³ = 8 (A1)

Answer: (a) 7 (b) 3 (c) 8

1. √50 = √(25 × 2) = 5√2 and √18 = √(9 × 2) = 3√2 (M1)
2. √8 = √(4 × 2) = 2√2, so the expression is 5√2 + 3√2 − 2√2 (M1)
3. 5√2 + 3√2 − 2√2 = 6√2 (A1)

Answer: 6√2

1. (a) Multiply top and bottom by √3: 1/√3 × √3/√3 = √3/3 (M1)
2. (a) = √3/3 (A1)
3. (b) Multiply top and bottom by √5: 10/√5 × √5/√5 = 10√5/5 (M1)
4. (b) 10√5/5 = 2√5 (A1)

Answer: (a) √3/3 (b) 2√5

1. (a) (2 + √3)² = 4 + 4√3 + 3 (M1)
2. (a) = 7 + 4√3 (A1)
3. (b) Multiply top and bottom by the conjugate 2 − √5: 3(2 − √5)/((2)² − (√5)²) (M1)
4. (b) The denominator is 4 − 5 = −1, so the fraction is −3(2 − √5) (M1)
5. (b) = 3√5 − 6 (A1)

Answer: (a) 7 + 4√3 (b) 3√5 − 6

1. (a) 8^2/3 = (8^1/3)² = 2² (M1)
2. (a) So k = 2 (A1)
3. (b) Multiply top and bottom by √5 + 1: (√5 + 1)²/((√5)² − 1²) (M1)
4. (b) = (5 + 2√5 + 1)/4 = (6 + 2√5)/4 (M1)
5. (b) Divide top and bottom by 2: (3 + √5)/2 (A1)

Answer: (a) k = 2 (b) (3 + √5)/2

1. Find two numbers with product 12 and sum −7: they are −3 and −4, so (x − 3)(x − 4) = 0 (M1)
2. x − 3 = 0 or x − 4 = 0, giving x = 3 or x = 4 (A1)

Answer: x = 3 or x = 4

1. Take out the common factor: 3x(x − 4) = 0 (M1)
2. 3x = 0 gives x = 0 (M1)
3. x − 4 = 0 gives x = 4, so the two roots are 0 and 4 (A1)

Answer: x = 0 or x = 4

1. Rearrange so one side is zero: x² − 5x − 24 = 0 (M1)
2. Factorise: (x − 8)(x + 3) = 0 (M1)
3. x = 8 or x = −3 (A1)

Answer: x = 8 or x = −3

1. Use x = (−b ± √(b² − 4ac)) ÷ 2a with a = 2, b = 4 and c = −1: x = (−4 ± √24) ÷ 4 (M1)
2. Simplify the surd: √24 = 2√6, so x = (−4 ± 2√6) ÷ 4 (M1)
3. Divide every term by 2: x = −1 ± (√6 ÷ 2) (A1)

Answer: x = −1 ± (√6 ÷ 2)

1. Set 12t − 5t² = 4 and rearrange: 5t² − 12t + 4 = 0 (M1)
2. Factorise: (5t − 2)(t − 2) = 0 (M1)
3. t = 2/5 or t = 2 (A1)

Answer: t = 2/5 s and t = 2 s

1. Let the width be x cm, so the length is x + 5 cm and x(x + 5) = 84 (M1)
2. Expand and factorise: x² + 5x − 84 = 0, so (x + 12)(x − 7) = 0 (M1)
3. x = 7, because x = −12 is impossible for a length (A1)
4. The rectangle is 7 cm by 12 cm (A1)

Answer: 7 cm by 12 cm

1. Use the formula with a = 4, b = −12 and c = 7: x = (12 ± √((−12)² − 4 × 4 × 7)) ÷ 8 (M1)
2. The discriminant is 144 − 112 = 32, and √32 = 4√2, so x = (12 ± 4√2) ÷ 8 (M1)
3. Divide every term by 4: x = (3 + √2) ÷ 2 (A1)
4. The second root is x = (3 − √2) ÷ 2 (A1)

Answer: x = (3 ± √2) ÷ 2

1. A 20% reduction leaves 80%, so £72 is 80% of the original price (M1)
2. Original price = 72 ÷ 0.8 (M1)
3. = £90.00 (A1)

Answer: £90.00

1. £477 is 106% of the old pay, so the multiplier is 1.06 (M1)
2. Old pay = 477 ÷ 1.06 (M1)
3. = £450.00 (A1)

Answer: £450.00

1. A 15% reduction leaves 85%, so £357 is 85% of the original price (M1)
2. Original price = 357 ÷ 0.85 (M1)
3. = £420.00 (A1)

Answer: £420.00

1. £486 is 120% of the price before VAT, so the multiplier is 1.2 (M1)
2. Price before VAT = 486 ÷ 1.2 (M1)
3. = £405.00 (A1)
4. VAT = 486 − 405 = £81.00 (A1)

Answer: Price before VAT £405.00, VAT £81.00

1. The selling price is 115% of the cost price, so the multiplier is 1.15 (M1)
2. Cost price = 138 ÷ 1.15 (M1)
3. = £120.00 (A1)
4. Profit = 138 − 120 = £18.00 (A1)

Answer: Cost price £120.00, profit £18.00

1. The two reductions give a single multiplier: 0.8 × 0.9 = 0.72 (M1)
2. Original price = 432 ÷ 0.72 (M1)
3. (a) = £600.00 (A1)
4. (b) The final price is 0.72 of the original, so the reduction is 1 − 0.72 = 0.28 (M1)
5. (b) = 28% (A1)

Answer: (a) £600.00 (b) 28%

1. The two changes give a single multiplier: 0.9 × 1.05 = 0.945 (M1)
2. Value before the changes = 15,120 ÷ 0.945 (M1)
3. (a) = £16,000.00 (A1)
4. (b) 0.945 = 1 − 0.055, so the value is 5.5% lower than at the start (M1)
5. (b) 5.5% decrease (A1)

Answer: (a) £16,000.00 (b) 5.5% decrease

1. The terms go up by 3 each time, so the nth term starts with 3n (M1)
2. Compare with 3n: 3, 6, 9, 12, ... and each term is 2 more than this (M1)
3. So the nth term is 3n + 2 (A1)

Answer: 3n + 2

1. (a) Substitute n = 15: 4 × 15 − 7 (M1)
2. (a) = 60 − 7 = 53 (A1)
3. (b) Solve 4n − 7 = 41, so 4n = 48 (M1)
4. (b) n = 12, so 41 is the 12th term (A1)

Answer: (a) 53 (b) the 12th term

1. Suppose 3n + 4 = 100 (M1)
2. 3n = 96, so n = 32 (M1)
3. 32 is a whole number, so yes: 100 is the 32nd term (A1)

Answer: Yes, 100 is the 32nd term

1. First differences are 5, 7, 9, 11 and the second difference is 2, so the nth term contains n² (M1)
2. Subtract n² = 1, 4, 9, 16, 25 from the terms: 1, 3, 5, 7, 9 (M1)
3. The remainder is the linear sequence 2n − 1 (A1)
4. So the nth term is n² + 2n − 1 (A1)

Answer: n² + 2n − 1

1. First differences are 7, 11, 15, 19 and the second difference is 4, so the nth term contains 2n² (M1)
2. Subtract 2n² = 2, 8, 18, 32, 50 from the terms: 1, 2, 3, 4, 5 (M1)
3. The remainder is the linear sequence n (A1)
4. So the nth term is 2n² + n (A1)

Answer: 2n² + n

1. (a) n² + 3n = 40, so n² + 3n − 40 = 0 (M1)
2. (a) (n + 8)(n − 5) = 0, so n = 5 (A1)
3. (b) n² + 3n = 88, so n² + 3n − 88 = 0 and (n + 11)(n − 8) = 0 (M1)
4. (b) n = 8, so 88 is the 8th term (A1)

Answer: (a) n = 5 (b) yes, 88 is the 8th term

1. First differences are 5, 7, 9 and the second difference is 2, so the nth term contains n² (M1)
2. Subtract n² = 1, 4, 9, 16 from the terms: 2, 4, 6, 8, so the remainder is 2n (M1)
3. (a) The nth term is n² + 2n (A1)
4. (b) Row 12: 12² + 2 × 12 = 144 + 24 = 168 (A1)
5. (c) n² + 2n = 143, so (n + 1)² = 144 and n = 11 (reject n = −13 because a row number must be positive) (A1)

Answer: (a) n² + 2n (b) 168 trees (c) row 11

1. Rearrange the second equation: x = y + 1 (M1)
2. Substitute into the first: 2(y + 1) + 3y = 12, so 5y = 10 and y = 2 (M1)
3. Then x = 2 + 1 = 3 (A1)

Answer: x = 3, y = 2

1. The y terms match, so subtract the second equation from the first: 2x = 4 (M1)
2. x = 2 (M1)
3. Substitute back: 2 + 2y = 12, so y = 5 (A1)

Answer: x = 2, y = 5

1. Let a coffee cost c and a tea cost t: 3c + 2t = 13 and 2c + 3t = 12 (M1)
2. Multiply the first equation by 3 and the second by 2: 9c + 6t = 39 and 4c + 6t = 24 (M1)
3. Subtract: 5c = 15, so c = 3 (M1)
4. Then 6 + 3t = 12, so t = 2 (A1)

Answer: A coffee costs £3 and a tea costs £2

1. Multiply the first equation by 2 and the second by 3: 8x + 6y = 50 and 9x + 6y = 54 (M1)
2. Subtract the first from the second: x = 4 (M1)
3. Substitute back: 16 + 3y = 25, so 3y = 9 (M1)
4. y = 3 (A1)

Answer: x = 4, y = 3

1. From the second equation y = 3x ÷ 2 (M1)
2. Substitute into the first: 2x + 3(3x ÷ 2) = 13, so 13x ÷ 2 = 13 (M1)
3. x = 2 (M1)
4. Then y = 6 ÷ 2 = 3 (A1)

Answer: x = 2, y = 3

1. Substitute y = x + 1 into the second equation: x² + (x + 1)² = 25 (M1)
2. Expand: 2x² + 2x + 1 = 25, so x² + x − 12 = 0 (M1)
3. Factorise: (x + 4)(x − 3) = 0 (M1)
4. x = 3 gives y = 4 (A1)
5. x = −4 gives y = −3 (A1)

Answer: x = 3, y = 4 and x = −4, y = −3

1. Let the sides be x cm and y cm: x + y = 13 and xy = 40 (M1)
2. Substitute y = 13 − x into the area equation: x(13 − x) = 40 (M1)
3. Rearrange and factorise: x² − 13x + 40 = 0, so (x − 5)(x − 8) = 0 (M1)
4. x = 5 or x = 8 (A1)
5. The sides are 5 cm and 8 cm, whichever way round they are taken (A1)

Answer: 5 cm by 8 cm