📈 Quadratics: factorise, complete the square, formula

Answer key

1. (c) x = 3 or x = 1/2 — Method: factorise into two brackets whose x terms multiply to 2x² and whose numbers multiply to 3, checking that they produce the middle term −7x, then set each bracket equal to zero. Working: (2x − 1)(x − 3) expands to 2x² − 6x − x + 3 = 2x² − 7x + 3, so (2x − 1)(x − 3) = 0; then 2x − 1 = 0 gives x = 1/2 and x − 3 = 0 gives x = 3. Answer: x = 3 or x = 1/2. The distractors: x = 3/2 or x = 1 comes from factorising as (2x − 3)(x − 1), whose middle term is −5x and not −7x; x = −3 or x = −1/2 comes from reading the roots straight out of (2x − 1)(x − 3) without changing the signs; x = 6 or x = 1 comes from using the quadratic formula with the denominator written as a instead of 2a, dividing 7 ± 5 by 2.
2. (d) (5x − 1)/(x² − x − 2) — The common denominator is (x + 1)(x − 2) = x² − x − 2. The numerator becomes 2(x − 2) + 3(x + 1) = (2x − 4) + (3x + 3) = 5x − 1, so the sum is (5x − 1)/(x² − x − 2). Choosing 5/(2x − 1) adds the numerators (2 + 3 = 5) and the denominators ((x + 1) + (x − 2) = 2x − 1) directly, which is not how fractions add. Choosing (5x + 1)/(x² − x − 2) comes from expanding 2(x − 2) as 2x − 2 instead of 2x − 4, losing part of the constant term. Choosing (5x − 1)/(x² + x − 2) has the correct numerator but the wrong sign on the x-term when (x + 1)(x − 2) is expanded.
3. (a) t = 0 or t = 4 — The ball is at ground level exactly when h = 0. From the table, h = 0 at t = 0 and at t = 4, so those are the two times. Distractor origins: t = 1 or t = 3 picks the times with equal (but non-zero) height instead of ground level; t = 2 picks the time of maximum height instead of ground level; t = 0 finds only the starting time and misses the second one.
4. (c) x = 3 and x = −5 — To factorise x² + 2x − 15, find two numbers that multiply to −15 and add to 2: these are 5 and −3, since 5 × (−3) = −15 and 5 + (−3) = 2. So x² + 2x − 15 = (x + 5)(x − 3). Setting each factor to zero gives x = −5 and x = 3. Choosing x = −3 and x = 5 comes from swapping the signs of the correct roots. Choosing x = 5 and x = 3 uses the right pair of numbers, 5 and 3, but forgets that one of them must be negative for the product to equal −15. Choosing x = −15 and x = 1 mistakes the constant term, −15, for one of the roots, and pairs it oddly with x = 1.
5. (b) x = 2, y = −5 — 2x² − 8x + 3 rewrites as 2(x² − 4x) + 3, then as 2[(x − 2)² − 4] + 3, which simplifies to 2(x − 2)² − 5, since −2 × 4 + 3 = −5. Substituting x = 2: 2 × 2² = 8, 8 × 2 = 16, so 8 − 16 + 3 = −5, confirming the minimum value −5 at x = 2: turning point x = 2, y = −5. Halving b instead of halving b/a — using 4 as the shift instead of 2 — lands on turning point x = 4, y = −29, which is wrong because when a ≠ 1 the shift inside the bracket is b/(2a), not b/a alone. Reading the bracket's sign directly as the turning point's x-coordinate gives x = −2, y = −5 — wrong, because (x − 2)² is zero at x = 2, not x = −2. Computing 8 − 3 = 5 instead of 3 − 8 = −5 flips the sign of the constant, giving x = 2, y = 5 — wrong, since the completed square's constant must be evaluated as 3 minus 8, not 8 minus 3. Whenever a ≠ 1, factor a out of the x² and x terms first, and always check a turning point by substituting back into the original equation.
6. (a) −10 — Method: multiply each bracket out, treating the second bracket as being multiplied by −3 because it is subtracted, then collect like terms. Working: 4(2x − 1) = 8x − 4 and −3(x + 2) = −3x − 6, so the expression becomes 8x − 4 − 3x − 6 − 5x; the x terms give 8x − 3x − 5x = 0, so no term in x survives, and the numbers give −4 − 6 = −10. Answer: −10. The distractors: 2 comes from expanding −3(x + 2) as −3x + 6, leaving the numbers −4 + 6; 5x − 10 comes from forgetting the final −5x, so the x terms give 8x − 3x = 5x; −7 comes from multiplying the 4 over only the first term of its bracket, giving 8x − 1 and so the numbers −1 − 6.
7. (a) 3x/2 — Factorise top and bottom first: 6x³ − 15x² = 3x²(2x − 5), and 4x² − 10x = 2x(2x − 5). The bracket (2x − 5) is common to both, so it cancels, leaving 3x²/(2x); dividing the power of x, 3x² ÷ x = 3x, gives 3x/2. Writing 3x²/2 cancels the (2x − 5) correctly and removes the denominator's x, but never reduces the power of x left in the numerator — 3x² ÷ x should give 3x, not stay as 3x². Writing 3/2 cancels the (2x − 5) correctly but then drops the x from the numerator altogether, treating 3x² over x as if it cancelled completely to 3 instead of reducing to 3x. Writing −3x/2 comes from factorising the denominator with the wrong sign, as 2x(5 − 2x) instead of 2x(2x − 5); cancelling (5 − 2x) against the numerator's (2x − 5) then needs an extra minus sign, which flips the answer to −3x/2.
8. (d) 12 m — Method: give the width a letter, write the length in terms of it and use length × width = area to form a quadratic. Working: with a width of x metres the length is x + 4, so x(x + 4) = 96, which rearranges to x² + 4x − 96 = 0; factorising gives (x + 12)(x − 8) = 0, and a width must be positive, so x = 8 and the length is 8 + 4 = 12. Answer: 12 m, and 12 × 8 = 96 as the area requires. The distractors: 8 m is the width rather than the length asked for; 16 m comes from taking the other root as 12 and adding 4 to it, ignoring that the root −12 cannot be a width; 24 m comes from dividing the area by the 4 in the question instead of forming an equation.
9. (a) x = 4 ± √13 — Method: halve the coefficient of x to form the bracket, subtract the square of that number to keep the expression equal to the original, then rearrange and take the square root of both sides. Working: half of −8 is −4, so x² − 8x + 3 = (x − 4)² − 16 + 3 = (x − 4)² − 13; setting this equal to zero gives (x − 4)² = 13, so x − 4 = ±√13 and x = 4 ± √13. Answer: x = 4 ± √13. The distractors: x = 8 ± √13 comes from putting the whole coefficient 8 inside the bracket instead of half of it; x = 4 ± √19 comes from adding 16 and 3 rather than subtracting 3 from 16; x = −4 ± √13 comes from writing the bracket as (x + 4)², which reverses the sign of the number that comes out of it.
10. (c) x = 5, y = 5 — x² − 10x + 30 = (x − 5)² − 5² + 30 = (x − 5)² + 5. Substituting x = 5: 5² = 25, 10 × 5 = 50, so 25 − 50 + 30 = 5, confirming the minimum value 5 at x = 5: turning point (5, 5). Using −10 instead of half of it inside the bracket gives (x − 10)² − 70, turning point (10, −70) — wrong, because only half the coefficient of x belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (−5, 5) — wrong, because (x − 5)² is zero at x = 5, not x = −5. Computing 25 − 30 = −5 instead of 30 − 25 = 5 flips the sign of the constant, giving (5, −5) — wrong, since the completed square's constant must be evaluated as 30 minus 25, not 25 minus 30. Always check a turning point by substituting its x-value back into the original equation.
11. (a) x = 4 or x = −3 — Method: find two numbers that multiply to give −12 and add to give −1 — these are −4 and 3. So x² − x − 12 = (x − 4)(x + 3) = 0, giving x = 4 or x = −3. Distractor origins: x = −4 or x = 3 has the signs the wrong way round; x = 4 or x = 3 makes both roots positive, ignoring the sign of −12; x = 12 or x = −1 comes from reading off the coefficient and the constant directly instead of factorising.
12. (d) x² − 49 — Area = length × width = (x + 7)(x − 7). Expanding: x × x = x², x × (−7) = −7x, 7 × x = 7x, 7 × (−7) = −49. The two middle terms −7x and 7x cancel, leaving x² − 49. A candidate who misremembers the difference-of-two-squares result as a sum gets x² + 49. A candidate who makes a sign error and treats both middle terms as −7x instead of cancelling gets x² − 14x − 49. A candidate who confuses area with perimeter and simply adds the length and width gets 2x.
13. (c) x = 2 or x = 3 — Method: factorise into two brackets whose numbers multiply to the constant term and add to the coefficient of x, then set each bracket equal to zero. Working: two numbers that multiply to 6 and add to −5 are −2 and −3, so x² − 5x + 6 = (x − 2)(x − 3) = 0; then x − 2 = 0 gives x = 2 and x − 3 = 0 gives x = 3. Answer: x = 2 or x = 3. The distractors: x = −2 or x = −3 comes from reading the numbers inside the brackets as the solutions instead of changing their signs; x = 1 or x = 6 comes from taking the first factor pair of 6 without checking that the pair adds to −5; x = 5 or x = 6 comes from reading the solutions straight off the 5 and the 6 in the equation.
14. (d) x = 3 or x = −3 — Method: get x² on its own with a coefficient of 1, then take the square root of both sides and keep both the positive and the negative root. Working: dividing 2x² = 18 by 2 gives x² = 9, and the square root of 9 is 3, so x = 3 or x = −3; both check, because 2 × 9 = 18 either way. Answer: x = 3 or x = −3. The distractors: x = 9 or x = −9 comes from dividing by 2 and then forgetting to take the square root; x = 4 or x = −4 comes from subtracting 2 from 18 instead of dividing by it, giving x² = 16; x = 6 or x = −6 comes from multiplying by 2 instead of dividing, giving x² = 36.
15. (b) x = −3, y = −7 — x² + 6x + 2 = (x + 3)² − 3² + 2 = (x + 3)² − 7. Substituting x = −3: (−3)² = 9, 6 × (−3) = −18, so 9 − 18 + 2 = −7, confirming the minimum value −7 at x = −3: turning point (−3, −7). Using 6 instead of half of 6 inside the bracket gives (x + 6)² − 34, turning point (−6, −34) — wrong, because only half the coefficient of x belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (3, −7) — wrong, because (x + 3)² is zero at x = −3, not x = 3. Computing 9 − 2 = 7 instead of 2 − 9 = −7 flips the sign of the constant, giving (−3, 7) — wrong, since the completed square's constant must be evaluated as 2 minus 9, not 9 minus 2. Always check a turning point by substituting its x-value back into the original equation.
16. (a) x = 6 or x = −4 — We need two numbers that multiply to −24 and add to −2: these are −6 and 4, since −6 × 4 = −24 and −6 + 4 = −2. So x² − 2x − 24 = (x − 6)(x + 4) = 0, giving x = 6 or x = −4. A candidate who swaps the signs, using 6 and −4 the wrong way round in the brackets, gets x = −6 or x = 4. A candidate who picks the wrong factor pair, 8 and −3 (which multiply to −24 but add to +5, not −2), gets (x + 8)(x − 3) = 0 and answers x = −8 or x = 3. A candidate who makes both factors negative gets x = −6 or x = −4, which would require the constant term to be +24, not −24.
17. (c) x = 0 or x = 3 — Method: take out the common factor 5x: 5x(x − 3) = 0, so 5x = 0 or x − 3 = 0, giving x = 0 or x = 3. Distractor origins: x = 0 or x = 15 comes from forgetting to divide the second term by the common factor 5x correctly, leaving 15 instead of 3; x = 3 loses the solution x = 0 by dividing both sides by x; x = 5 or x = 3 mistakes the coefficient 5 itself for one of the solutions.
18. (d) (3x − 2)(x + 4) — For 3x² + 10x − 8, find two numbers multiplying to 3 × (−8) = −24 and adding to 10: these are 12 and −2. Rewrite: 3x² + 12x − 2x − 8 = 3x(x + 4) − 2(x + 4) = (3x − 2)(x + 4). Choosing (3x + 2)(x − 4) expands to 3x² − 10x − 8 — the correct factor pair but the wrong signs, giving the middle term the wrong sign. Choosing (x − 2)(3x + 4) expands to 3x² − 2x − 8 — the 3 is attached to the wrong bracket, changing which terms combine for the x-coefficient. Choosing (3x − 4)(x + 2) expands to 3x² + 2x − 8 — this uses 4 and 2 instead of the correct pair 12 and 2, so the middle term does not come to 10x.
19. (b) (2x + 14)/((x − 1)(x + 3)) — To subtract these fractions, write them over the common denominator (x − 1)(x + 3): the numerator becomes 4(x + 3) − 2(x − 1). Expanding gives 4x + 12 − 2x + 2, which simplifies to 2x + 14, so the answer is (2x + 14)/((x − 1)(x + 3)). Writing (2x + 11)/((x − 1)(x + 3)) comes from not distributing the minus sign fully across the second bracket, treating −2(x − 1) as −2x − 1 instead of −2x + 2. Writing 2/((x − 1)(x + 3)) comes from subtracting the two original numerators directly, 4 − 2 = 2, the same mistake as subtracting fractions without a common denominator. Writing (2x − 10)/((x − 1)(x + 3)) comes from swapping which numerator multiplies which bracket, forming 4(x − 1) − 2(x + 3) instead of 4(x + 3) − 2(x − 1).
20. (d) x = 2 or x = −5 — Method: find two numbers that multiply to give −10 and add to give 3 — these are 5 and −2. So x² + 3x − 10 = (x + 5)(x − 2) = 0, giving x = −5 or x = 2. Distractor origins: x = −2 or x = 5 swaps the signs of the two roots; x = 2 or x = 5 makes both roots positive, ignoring the sign of −10; x = −5 or x = −2 makes both roots negative.
21. (b) 5 and −6 — Method: turn the sentence into an equation in one letter, collect every term on one side so it reads as a quadratic equal to zero, then factorise. Working: if the number is x then x² + x = 30, which rearranges to x² + x − 30 = 0; two numbers that multiply to −30 and add to 1 are 6 and −5, so (x + 6)(x − 5) = 0 and x = −6 or x = 5. Both values work: 5 + 25 = 30 and −6 + 36 = 30. Answer: 5 and −6. The distractors: 6 and −5 comes from reading the numbers inside the brackets as the solutions without changing their signs; 5 only comes from discarding the negative solution, although nothing in the question rules out a negative number; 5 and 6 comes from hunting for a factor pair of 30 instead of forming and solving the quadratic.
22. (a) y = (x − 2)² + 3 — Since (x − 2)² is never negative, (x − 2)² + 3 is always at least 3, so y can never equal 0 and the graph never crosses the x-axis. The other three graphs are all given in a factorised or difference-of-squares form that shows two real roots: y = (x − 2)(x + 3) crosses at x = 2 and x = −3; y = x² − 9 = (x − 3)(x + 3) crosses at x = 3 and x = −3; y = (x + 4)(x − 1) crosses at x = −4 and x = 1.
23. (c) x = 8 — The turning point lies exactly halfway between the two roots. If the other root is r, the midpoint of −2 and r must be 3, so (−2 + r) ÷ 2 = 3, giving r = 8. Choosing x = 5 comes from adding 2 and 3 rather than using the midpoint relationship correctly. Choosing x = 1 comes from subtracting 2 from 3 instead of reflecting −2 across the turning point. Choosing x = −8 finds the right distance but then reflects in the y-axis instead of in the line of symmetry x = 3, so the sign of the answer is flipped.
24. (d) 3x² + 4x − 7 — Method: the minus sign in front of the second bracket changes the sign of every term inside it; then collect like terms. Working: removing the brackets gives 5x² + 3x − 2 − 2x² + x − 5; the squared terms give 5x² − 2x² = 3x², the x terms give 3x + x = 4x, and the number terms give −2 − 5 = −7. Answer: 3x² + 4x − 7. The distractors: 7x² + 2x + 3 comes from adding the two brackets instead of subtracting, giving 5x² + 2x², 3x − x and −2 + 5; 3x² + 2x + 3 comes from applying the minus sign to 2x² only, leaving −x and +5 unchanged so that 3x − x = 2x and −2 + 5 = 3; 3x² + 4x + 3 comes from changing the signs of the terms with letters but leaving +5 as it stood, so the number terms give −2 + 5 = 3.