📈 Quadratics: factorise, complete the square, formula

Answer key

1. (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.
2. (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.
3. (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.
4. (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.
5. (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.
6. (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.
7. (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.
8. (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).
9. (d) 3 — 2x² − 12x + 7 rewrites as 2(x² − 6x) + 7, then as 2[(x − 3)² − 9] + 7, which simplifies to 2(x − 3)² − 11, since −2 × 9 + 7 = −11. The bracket (x − 3)² is zero when x = 3, so the minimum occurs at x = 3. Treating the shift as b/a instead of b/(2a) — using 6 instead of 3 — gives x = 6, which is wrong. Reading the bracket's sign directly without negating it gives x = −3, wrong, because (x − 3)² is zero at x = 3, not x = −3. Reading off the coefficient of x itself, −12, and calling that the answer skips the completing-the-square process entirely and gives x = −12, which is wrong because b is not the turning point's x-coordinate under any circumstance. Always check: substituting your value of x should make the bracketed term equal to zero, and nothing else.
10. (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.
11. (b) (3x + 1)(2x − 3) — To factorise 6x² − 7x − 3, look for two numbers that multiply to 6 × (−3) = −18 and add to −7: these are −9 and 2. Splitting the middle term gives 6x² − 9x + 2x − 3, which groups to 3x(2x − 3) + 1(2x − 3) = (3x + 1)(2x − 3). Writing (3x − 1)(2x + 3) comes from using the pair 9 and −2 instead — the right product but the wrong signs, giving +7x instead of −7x. Writing (6x − 1)(x + 3) comes from picking the factor pair 18 and −1: it multiplies to −18 correctly, but it adds to 17, not −7, so checking only the product splits the middle term as 6x² + 18x − x − 3 and pairs the wrong factors of 6 and 3 together. Writing (x − 3)(6x + 1) comes from that same unchecked pair used the other way round, splitting the middle term as 6x² + x − 18x − 3.
12. (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.
13. (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.
14. (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.
15. (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.
16. (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.
17. (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.
18. (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.
19. (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.
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. (b) (x − 3)/(x + 2) — Factorise both: x² − 9 = (x − 3)(x + 3) (difference of two squares), and x² + 5x + 6 = (x + 2)(x + 3) (two numbers multiplying to 6 and adding to 5, namely 2 and 3). The factor (x + 3) is common to both, so it cancels, leaving (x − 3)/(x + 2). Choosing −9/(5x + 6) comes from cancelling the x² terms directly without factorising first — x² is not a common factor of the whole numerator or denominator. Choosing (x + 3)/(x + 2) cancels the (x − 3) factor instead of the shared (x + 3) factor, and (x − 3) does not appear in the denominator to cancel with. Choosing x − 3 cancels the whole denominator (x + 2) as though it were equal to 1.
22. (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.
23. (b) 6x + 6 — Perimeter = 2[(2x + 5) + (x − 2)] = 2(3x + 3) = 6x + 6. A candidate who adds the length and width but forgets to double for the perimeter gets 3x + 3. A candidate who makes a sign error and adds 2 instead of subtracting it before doubling gets 2[(2x + 5) + (x + 2)] = 6x + 14. A candidate who multiplies the length and width instead of adding them, confusing the perimeter formula with the area formula, and then doubles that product, gets 2(2x + 5)(x − 2) = 4x² + 2x − 20.
24. (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.