📈 Quadratics: factorise, complete the square, formula

Answer key

1. (d) t = 5 — The maximum height occurs halfway between the two times when the ball is at ground level: the midpoint of t = 1 and t = 9 is (1 + 9) ÷ 2 = 5, so the ball reaches its maximum height at t = 5 seconds. Getting t = 4 comes from halving the DIFFERENCE between the times, 9 − 1 = 8, then 8 ÷ 2 = 4, instead of finding their midpoint. Getting t = 8 uses that difference, 9 − 1 = 8, as if the gap between the two times were itself the time of the maximum. Getting t = 10 adds the two times, 1 + 9 = 10, but forgets to divide by 2.
2. (d) 2x/(x − 2) — Factorise x² − 4 as (x − 2)(x + 2) first — it's a difference of two squares. The (x + 2) in the numerator then cancels with the (x + 2) in the factorised denominator, and 6x² ÷ 3x simplifies to 2x, leaving 2x/(x − 2). Writing 2/(x − 2) comes from over-cancelling 6x² ÷ 3x as 2 instead of 2x, dropping the x that should remain. Writing 2x/(x + 2) comes from factorising x² − 4 as (x + 2)² instead of (x − 2)(x + 2), a difference of two squares always has one plus and one minus bracket. Writing 2x²/(x − 2) comes from simplifying 6x² ÷ 3x as 2x² instead of 2x, dividing the coefficients but not reducing the power of x.
3. (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.
4. (d) (x − 7)² + 1 is at least 1 for every x, so y is never 0 — x² − 14x + 50 = (x − 7)² − 7² + 50 = (x − 7)² + 1. Since a squared term can never be negative, (x − 7)² ≥ 0 for every real x, so y = (x − 7)² + 1 ≥ 1 for every x — y is always at least 1, so it can never equal 0, and the curve never crosses the x-axis. Claiming the discriminant is positive is wrong on the facts: b² − 4ac = (−14)² − 4(1)(50) = 196 − 200 = −4, which is NEGATIVE, not positive — a negative discriminant is in fact the standard reason a quadratic has no real roots, so this option misstates its own sign. Claiming the turning point is at (7, −1) is wrong because completing the square gives (x − 7)² + 1, so the turning point is (7, 1) — the constant is +1, not −1, and a point with a negative y-coordinate would sit BELOW the x-axis, not above it, contradicting the option's own claim. Claiming that 50 being positive is enough is wrong because it ignores the −14x term completely — a positive constant term alone does not guarantee y stays positive for every x (the expression still dips down towards its turning point before growing again, and it is the completed square that shows how far it dips, not the constant on its own).
5. (c) x = 0 or x = −7 — Factorising: x² + 7x = x(x + 7) = 0, so x = 0 or x + 7 = 0, giving x = 0 or x = −7. A candidate who divides both sides of the original equation by x, which loses the solution x = 0, gets only x = −7. A candidate who makes a sign error solving x + 7 = 0 gets x = 0 or x = 7. A candidate who misreads the coefficient and doubles it gets x = 0 or x = −14.
6. (c) 12 cm — For a square, area = side². So side² = 144, giving side = ±12. Since a length must be positive, the side length is 12 cm. A candidate who gives both square roots without rejecting the negative one, which cannot be a length, answers 12 cm or −12 cm. A candidate who halves 144 instead of taking its square root gets 72 cm. A candidate who divides 144 by 4, confusing the area formula with a perimeter calculation, gets 36 cm.
7. (c) t = 1 second and t = 5 seconds — The ball is at ground level when h = 0, which happens when either bracket is zero. t − 1 = 0 gives t = 1, and 5 − t = 0 gives t = 5. The option with t = −1 makes a sign error solving the first bracket. The option with t = −5 makes a sign error solving the second bracket, treating 5 − t = 0 as if it gave a negative solution. The option with t = 4 seconds is an arithmetic slip in solving 5 − t = 0.
8. (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.
9. (d) x² − 6x + 9 — Method: squaring a bracket means multiplying that bracket by itself, so expand (x − 3)(x − 3) term by term and then collect like terms. Working: x × x = x², x × (−3) = −3x, (−3) × x = −3x and (−3) × (−3) = 9, giving x² − 3x − 3x + 9, and the two middle terms collect to −6x. Answer: x² − 6x + 9. The distractors: x² − 3x + 9 comes from writing down only one of the two middle products instead of both; x² + 6x + 9 comes from treating (−3) × x as +3x, so the middle terms are added rather than subtracted; x² − 9 comes from treating the square as the difference of two squares (x − 3)(x + 3).
10. (c) 25 — Method: read off a, b and c with their signs and substitute them into b² − 4ac. Working: for 2x² + 3x − 2 = 0, a = 2, b = 3 and c = −2, so b² − 4ac = 3² − 4 × 2 × (−2) = 9 − (−16) = 9 + 16 = 25. Answer: 25. The distractors: −7 comes from taking c as +2, which gives 9 − 16; 22 comes from working b² as 2 × 3 = 6 and then 6 + 16; 13 comes from leaving the 4 out of 4ac and working 9 − 2 × (−2).
11. (d) 8(7x − 3) — Method: find the highest common factor of the two terms, write it in front of a bracket and divide each term by it. Working: 56 = 8 × 7 and 24 = 8 × 3, so the highest common factor is 8; dividing gives 56x ÷ 8 = 7x and 24 ÷ 8 = 3, and the subtraction sign stays between them. Answer: 8(7x − 3), which multiplies back out to 56x − 24. The distractors: 8(7x + 3) comes from dropping the minus sign of −24 while dividing; 8(56x − 3) comes from dividing only the number term by 8 and leaving 56x untouched inside the bracket; 8(7x − 16) comes from subtracting 8 from 24 instead of dividing 24 by 8.
12. (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.
13. (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.
14. (d) It crosses the x-axis at x = 3 and x = −3. — y = x² − 9 factorises as (x − 3)(x + 3), since 9 = 3², so the curve crosses the x-axis at x = 3 and x = −3. Saying it crosses once at x = 9 mistakes the constant term for a root directly, without taking its square root. Saying it crosses at x = 9 and x = −9 makes the same mistake but adds a sign either way. Saying it does not cross the x-axis confuses the y-intercept, which is negative at (0, −9), with the number of times the curve meets the x-axis — a negative y-intercept combined with an upward-opening curve guarantees it crosses the x-axis twice.
15. (d) a/2 — Method: write both terms over the same denominator, subtract the numerators, then cancel the fraction down. Working: 6 is a multiple of 3, so a/3 is rewritten as 2a/6; the calculation becomes 5a/6 − 2a/6 = 3a/6, and dividing numerator and denominator by 3 gives a/2. Answer: a/2. The distractors: 4a/3 comes from subtracting the denominators as well as the numerators, giving (5 − 1)a over (6 − 3); 2a/3 comes from subtracting 1 from 5 without first rewriting a/3 as 2a/6, giving 4a/6; 7a/6 comes from adding the two fractions instead of subtracting them, giving 5a/6 + 2a/6.
16. (d) x = −2, y = −5 — 3x² + 12x + 7 rewrites as 3(x² + 4x) + 7, then as 3[(x + 2)² − 4] + 7, which simplifies to 3(x + 2)² − 5, since −3 × 4 + 7 = −5. Substituting x = −2: 3 × (−2)² = 12, 12 × (−2) = −24, so 12 − 24 + 7 = −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 = −41, 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 12 − 7 = 5 instead of 7 − 12 = −5 flips the sign of the constant, giving x = −2, y = 5 — wrong, since the completed square's constant must be evaluated as 7 minus 12, not 12 minus 7. 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.
17. (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.
18. (c) 6x(2x + 3) — The highest common factor of 12x² and 18x is 6x. Dividing each term by 6x gives 12x² ÷ 6x = 2x and 18x ÷ 6x = 3, so 12x² + 18x = 6x(2x + 3). A candidate who only takes out the number 6 (missing the x) gets 6(2x² + 3x), which is not fully factorised. A candidate who only takes out 2x (missing the extra factor of 3 in 6) gets 2x(6x + 9), also not fully factorised — the bracket still shares a common factor. A candidate who takes out 3x instead of the full 6x gets 3x(4x + 6), which again is not fully factorised since 4x + 6 shares a common factor of 2.
19. (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.
20. (a) It touches the x-axis once, only at x = 4. — (x − 4)² is a square, so it equals zero only when x − 4 = 0, that is at x = 4 — the curve just touches the x-axis there rather than crossing it, since a square cannot be negative on either side to cross through. Saying it crosses at x = 4 and x = −4 wrongly introduces a plus-or-minus, as if taking a square root of x, rather than recognising the bracket is already squared and only zero once. Saying it never touches the x-axis forgets that a squared term CAN equal zero, even though it can never be negative. Saying it crosses at x = 2 and x = −2 confuses (x − 4)² with the different expression x² − 4.
21. (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.
22. (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.
23. (b) x = 5 — Area = length × width, so x(x + 3) = 40, which rearranges to x² + 3x − 40 = 0. This factorises as (x + 8)(x − 5) = 0: the two numbers in the brackets must multiply to −40 and add to +3, and the pair 8 and −5 does both. This gives x = −8 or x = 5. Since x is a length, it cannot be negative, so x = 5. A candidate who gives both solutions without rejecting the negative one, which cannot be a length, answers x = 5 or x = −8. A candidate who picks the wrong factor pair of 40, such as 10 and −4 instead of 8 and −5, gets (x + 10)(x − 4) = 0 and answers x = 4. A candidate who rejects the wrong root, keeping the negative solution instead of the positive one, answers x = −8.
24. (d) x = 3 — The table is symmetrical about the turning point: y = 0 at both x = 1 and x = 5, and the lowest value, y = −4, occurs exactly halfway between them, at x = 3. Choosing x = 5 picks one of the roots rather than the midpoint between them. Choosing x = 1 picks the other root for the same reason. Choosing x = 6 picks the x-value where y returns to its starting value of 5, which is not the turning point.