📈 Quadratics: factorise, complete the square, formula

Answer key

1. (a) k = 3 — Method: a solution of an equation makes both sides balance, so substitute it in and solve the equation in k that is left. Working: putting x = 3 gives 3² − (k + 1) × 3 + k = 0, that is 9 − 3k − 3 + k = 0, so 6 − 2k = 0 and k = 3; the equation is then x² − 4x + 3 = 0, whose solutions are 3 and 1. Answer: k = 3. The distractors: k = −3 comes from solving 6 − 2k = 0 as though it gave 2k = −6; k = 4 comes from expanding −3(k + 1) as −3k − 1, multiplying only the k by 3; k = 1.5 comes from working 3² as 3 × 2 = 6, which leaves 3 − 2k = 0.
2. (c) 27 days — The campaign starts at d = 3 and ends at d = 30, so it runs for 30 − 3 = 27 days. Getting 33 days comes from adding the two values, 3 + 30 = 33, instead of subtracting them. Getting 30 days uses only the end day and ignores that the campaign did not start at day 0. Getting 24 days comes from subtracting the start day twice, 30 − 3 − 3 = 24, instead of once.
3. (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.
4. (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.
5. (d) x³ + 6x² + 11x + 6 — (x + 1)(x + 2) = x² + 3x + 2. Multiplying by (x + 3): (x² + 3x + 2)(x + 3) = x³ + 3x² + 2x + 3x² + 9x + 6, which simplifies to x³ + 6x² + 11x + 6. Choosing x³ + 6x² + 6x + 6 has the right x² and constant terms but adds 1 + 2 + 3 = 6 for the x-coefficient instead of the correct sum of pairwise products 1×2 + 1×3 + 2×3 = 11. Choosing x³ + 5x² + 11x + 6 sums only two of the three constants (2 + 3 = 5) for the x² coefficient, leaving out the 1. Choosing x³ + 6x² + 11x + 5 adds the last two constants (2 + 3 = 5) instead of multiplying all three (1 × 2 × 3 = 6) for the constant term.
6. (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.
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. (c) x = 2 and x = 3 — x² − 5x + 6 factorises as (x − 2)(x − 3), since −2 and −3 multiply to give 6 and add to give −5. The graph crosses the x-axis where each bracket equals zero: x − 2 = 0 gives x = 2, and x − 3 = 0 gives x = 3. The option x = −2 and x = −3 comes from reading the signs inside the brackets directly instead of solving x − 2 = 0 and x − 3 = 0. The option x = 2 and x = −3 mixes up the sign of only one root. The option x = −1 and x = −6 comes from picking the wrong pair of factors of 6 (1 and 6 instead of 2 and 3) and then reading their signs directly from the brackets.
9. (a) x = 3 — Method: factorise x² + 2x − 15 as (x + 5)(x − 3), since 5 × (−3) = −15 and 5 + (−3) = 2. Setting each bracket equal to zero gives x + 5 = 0 or x − 3 = 0, so x = −5 or x = 3. Only x = 3 is offered here. Distractor origins: x = −3 reverses the sign of the factor pair, treating the bracket (x − 3) as giving x = −3 instead of x = 3; x = 5 takes the number from the other factor, (x + 5), but with the wrong sign, giving x = 5 instead of x = −5; x = 15 takes the constant term of the original expression as if it were a root, without factorising at all.
10. (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.
11. (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.
12. (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.
13. (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.
14. (c) It has two different real solutions — Method: the discriminant is the quantity under the square root sign in the quadratic formula, so its sign decides how many real values the formula produces. Working: when the discriminant is positive its square root is a real number that is not zero, and the formula adds that root and subtracts it in turn, so the two results are different; a positive discriminant that is not a perfect square still gives two solutions, but they are not whole numbers. Answer: it has two different real solutions. The distractors: one repeated real solution is what a discriminant of zero gives, because adding and subtracting zero changes nothing; no real solutions is what a negative discriminant gives, because a negative number has no real square root; two whole-number solutions happens only when the discriminant is a perfect square and the division works out exactly, which a positive discriminant does not guarantee.
15. (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.
16. (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.
17. (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.
18. (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.
19. (d) x = −4, y = −19 — x² + 8x − 3 = (x + 4)² − 4² − 3 = (x + 4)² − 19. Substituting x = −4: (−4)² = 16, 8 × (−4) = −32, so 16 − 32 − 3 = −19, confirming the minimum value −19 at x = −4: turning point (−4, −19). Using 8 instead of half of it inside the bracket gives (x + 8)² − 67, turning point (−8, −67) — 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 (4, −19) — wrong, because (x + 4)² is zero at x = −4, not x = 4. Computing 16 + 3 = 19 instead of −3 − 16 = −19 flips the sign of the constant, giving (−4, 19) — wrong, since the completed square's constant must be evaluated as −3 minus 16, not 16 plus 3. Check by substitution whenever the sign of a constant feels uncertain.
20. (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.
21. (d) x = 0 or x = 6 — Method: factorise by taking out the common factor x: x(x − 6) = 0, so x = 0 or x − 6 = 0, giving x = 0 or x = 6. Distractor origins: x = 6 comes from dividing both sides by x, which loses the solution x = 0; x = 0 stops after finding only one factor; x = 3 comes from halving the coefficient of x instead of factorising.
22. (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.
23. (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.
24. (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.