📈 Quadratics: factorise, complete the square, formula

Answer key

1. (b) −23 — Expand two of the three brackets first: (x + 5)(x − 2) = x² + 3x − 10. Then multiply this by the remaining bracket: (2x − 1)(x² + 3x − 10) = 2x³ + 6x² − 20x − x² − 3x + 10, which simplifies to 2x³ + 5x² − 23x + 10, so the coefficient of x is −23. Writing −13 comes from a sign slip in the first expansion, combining 5x − 2x as −5x − 2x = −7x instead of +3x, which carries through to a wrong final coefficient. Writing −20 comes from forgetting to distribute the −1 across every term of x² + 3x − 10, dropping the −1 × 3x = −3x contribution. Writing 10 comes from reading off the constant term of the expansion instead of the coefficient of x.
2. (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.
3. (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.
4. (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.
5. (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.
6. (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.
7. (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).
8. (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.
9. (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.
10. (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.
11. (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.
12. (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.
13. (d) If the graph has two real roots, they are equal and opposite in value, so they sum to zero. — A turning point on the y-axis means the graph's axis of symmetry is the line x = 0, so any two roots must be symmetrical about x = 0 — equal in size but opposite in sign, summing to zero. The graph could still have no real roots, but that is not guaranteed just from the turning point's position, so the option claiming it must have none is too strong. The roots do not have to both be positive — if real, one is positive and one negative (or both are zero). The graph does not have to touch the x-axis at exactly one point either; it could cross at two symmetrical points, touch at one point, or miss the x-axis entirely.
14. (d) 26 — Method: expand both brackets, treating the second as multiplication by −2 so that both of its terms change sign, then collect like terms. Working: 2(a + 6) = 2a + 12 and −2(a − 7) = −2a + 14, so the expression becomes 2a + 12 − 2a + 14; the a terms give 2a − 2a = 0, so no term in a survives, and the numbers give 12 + 14 = 26. Answer: 26. The distractors: −2 comes from expanding the second bracket as −2a − 14, so that the numbers give 12 − 14; 4a − 2 comes from adding 2(a − 7) instead of subtracting it, giving 2a + 12 + 2a − 14; 13 comes from multiplying the 2 over only the first term of each bracket, giving 2a + 6 − 2a + 7.
15. (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.
16. (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.
17. (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.
18. (a) x = 10 or x = −10 — Rearranging, x² = 100. Taking the square root of both sides gives x = ±10, i.e. x = 10 or x = −10. A candidate who forgets the negative root gives only x = 10. A candidate who halves 100 instead of taking its square root gets x = 50. A candidate who applies the ± sign to 100 itself instead of to its square root gets x = 100 or x = −100.
19. (d) y = (x + 3)(x − 5) — A root at x = −3 means the matching bracket must be zero when x = −3, so the bracket is (x − (−3)) = (x + 3). A root at x = 5 means the other bracket is (x − 5). So the equation is y = (x + 3)(x − 5); checking the y-intercept, (0 + 3)(0 − 5) = 3 × (−5) = −15, which matches the given value. The option (x − 3)(x + 5) swaps the signs of both roots. The option (x + 3)(x + 5) keeps the correct sign for the first root but gets the second wrong. The option (x − 3)(x − 5) gets the first root's sign wrong.
20. (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.
21. (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.
22. (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).
23. (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.
24. (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.