📈 Quadratics: factorise, complete the square, formula

Answer key

1. (a) x ≈ 6.70 — Since the two roots sum to 7, the other root is 7 − 0.30 = 6.70. The option 7.30 comes from adding the given root to 7 instead of subtracting it. The option 6.30 comes from subtracting 0.70 (one minus the given root) rather than the given root itself. The option 0.70 confuses the required root with the amount by which the given root falls short of 1.
2. (b) (2x + 1)/(x − 3) — Factorise both the numerator and the denominator before you cancel anything. The numerator 2x² + 7x + 3 factorises to (2x + 1)(x + 3), and the denominator x² − 9 is a difference of two squares, factorising to (x − 3)(x + 3). The (x + 3) factor is common to both, so it cancels, leaving (2x + 1)/(x − 3). Writing (2x + 1)/(x + 3) comes from factorising x² − 9 as (x + 3)² instead of (x − 3)(x + 3) — a difference of two squares always has one plus and one minus bracket. Writing (2x + 3)/(x − 3) comes from mis-factorising the numerator as (2x + 3)(x + 1) and then wrongly cancelling the (x + 1) against the denominator's (x + 3) as though they were the same bracket. Writing 2x + 1 comes from cancelling the (x + 3) factor correctly but then dropping the remaining (x − 3) on the denominator altogether.
3. (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.
4. (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.
5. (b) (x − 5)(x + 2) — We need two numbers that multiply to −10 and add to −3: these are −5 and 2, since −5 × 2 = −10 and −5 + 2 = −3. So x² − 3x − 10 = (x − 5)(x + 2). A candidate who swaps the signs, using +5 and −2, gets (x + 5)(x − 2), which expands to x² + 3x − 10 — the wrong middle term. A candidate who picks the factor pair 1 and 10 instead of 2 and 5 gets (x − 10)(x + 1), which expands to x² − 9x − 10. A candidate who makes both factors negative gets (x − 5)(x − 2), which expands to x² − 7x + 10 — the wrong sign on the constant term.
6. (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.
7. (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).
8. (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.
9. (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.
10. (c) 5(x + y − 1) — Method: take out the highest common factor of all three terms and divide every term by it, the number term included. Working: the highest common factor of 5x, 5y and −5 is 5; dividing gives 5x ÷ 5 = x, 5y ÷ 5 = y and −5 ÷ 5 = −1, so the bracket holds x + y − 1. Answer: 5(x + y − 1), which multiplies back out to 5x + 5y − 5. The distractors: 5(x + y + 1) comes from dividing −5 by 5 and losing the minus sign; 5(x + y − 5) comes from dividing only the terms containing a letter by 5 and carrying the −5 into the bracket unchanged; 5(xy − 1) comes from collecting the unlike terms 5x and 5y as 5xy before factorising.
11. (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.
12. (d) (x − 8)(x + 8) — x² − 64 = x² − 8², a difference of two squares, which factorises as (x − 8)(x + 8). A candidate who treats it as a perfect square with a repeated negative factor gets (x − 8)(x − 8), which expands to x² − 16x + 64 — wrong on both the middle and constant terms. A candidate who uses a repeated positive factor gets (x + 8)(x + 8), which expands to x² + 16x + 64. A candidate who picks a different factor pair of 64, such as 4 and 16, without checking that the middle term cancels, gets (x − 4)(x + 16), which expands to x² + 12x − 64 — the wrong middle term.
13. (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.
14. (d) k < 4 — Method: the number of real solutions of ax² + bx + c = 0 is decided by the discriminant b² − 4ac, and two different real solutions need it to be greater than zero. Working: here a = 1, b = −4 and c = k, so b² − 4ac = 16 − 4k; the condition is 16 − 4k > 0, which gives 16 > 4k and then k < 4. Answer: k < 4; for example k = 3 gives x² − 4x + 3 = 0, whose solutions are 1 and 3. The distractors: k ≤ 4 comes from using b² − 4ac ≥ 0, which also allows the single repeated solution at k = 4; k > 4 comes from dividing −4k > −16 by −4 without reversing the inequality sign; k < 16 comes from leaving the factor 4 out of 4ac and solving 16 − k > 0.
15. (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.
16. (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.
17. (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.
18. (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.
19. (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.
20. (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.
21. (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.
22. (a) x = 3/2 or x = −3/2 — Method: the equation has an x² term and a number but no x term, so make x² the subject and then take the square root of both sides, keeping the negative root as well as the positive one. Working: adding 9 to both sides of 4x² − 9 = 0 gives 4x² = 9, and dividing both sides by 4 gives x² = 9/4. Square-rooting the top and the bottom of 9/4 gives 3/2, so x = 3/2 or x = −3/2, and each value checks out because 4 × 9/4 − 9 = 0. Answer: x = 3/2 or x = −3/2. The distractors: x = 3 or x = −3 comes from square-rooting both sides of 4x² = 9 without first dividing by the 4, so the coefficient of x² is ignored; x = 9/4 or x = −9/4 comes from stopping at x² = 9/4 and writing that value down as x, leaving the square root undone; x = 3/2 only comes from taking the positive square root of 9/4 and losing the negative solution.
23. (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.
24. (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.