Printable · GCSE Higher · ages 14-16
Linear and quadratic inequalities worksheet — GCSE Higher
Fifteen questions on "linear and quadratic inequalities" — DfE statement A22. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
part Higher
Answer key: Linear and quadratic inequalities worksheet — GCSE Higher
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- (c) every value of x except x = −3 — Method: factorise the quadratic, then use the fact that a squared bracket is never negative to decide where the expression is strictly greater than zero. Working: x² + 6x + 9 factorises as (x + 3)², and a square is greater than or equal to 0 for every value of x; (x + 3)² is equal to 0 only when x + 3 = 0, that is when x = −3, so it is strictly greater than 0 at every other value. Answer: every value of x except x = −3. The distractors: 'every value of x, with no exceptions' comes from remembering that a square cannot be negative but forgetting that it can be zero, which a strict > rules out; 'x > −3 only' comes from taking the square root of both sides to get x + 3 > 0 and keeping only that branch; 'x < −3 only' comes from the same square-rooting followed by turning the sign round, as though the step had been a division by a negative number.
- (b) x ≤ 4 — Method: undo the number subtracted from 3x first, then divide by 3; the direction of the sign changes only for division by a negative number. Working: adding 1 to both sides of 3x − 1 ≤ 11 gives 3x ≤ 12; dividing both sides by 3, which is positive, gives x ≤ 4. Answer: x ≤ 4. The distractors: x ≥ 4 comes from turning the sign round while dividing by 3; x ≤ 10/3 comes from subtracting 1 from both sides instead of adding it, giving 3x ≤ 10; x ≤ 12 comes from stopping at 3x ≤ 12 and reading off the 12 without dividing by 3.
- (b) 9 — Subtract 14 from both sides: 4s ≤ 36. Divide both sides by 4: s ≤ 9, so the greatest number of tickets is 9. A candidate who forgets the £14 coach cost solves 4s ≤ 50, getting s ≤ 12.5, rounded down to 12. A candidate who adds the £14 instead of subtracting it solves 4s ≤ 64, getting s = 16. A candidate who miscalculates 50 − 14 as 32 solves 4s ≤ 32, getting s = 8.
- (d) Below y = x + 1, below x + y = 5, above y = 0 — For y ≤ x + 1, R lies on or below the line y = x + 1. For x + y ≤ 5 (that is, y ≤ 5 − x), R lies on or below that line too. For y ≥ 0, R lies on or above the x-axis. Combining all three: R is below y = x + 1, below x + y = 5, and above y = 0. Distractor routes: "Above y = x + 1" flips the first inequality, describing the wrong side of that line. "Above x + y = 5" flips the second inequality, describing the wrong side of that line. "Below y = 0" flips the third inequality, describing the wrong side of the x-axis.
- (b) −2, −1, 0, 1, 2, 3, 4 — Factorise x² − 2x − 8 = (x − 4)(x + 2), giving roots x = 4 and x = −2. Since the coefficient of x² is positive and the inequality is ≤ 0, the solution is the closed interval between the roots, −2 ≤ x ≤ 4, with the roots included because the inequality is not strict. The integers in this interval are −2, −1, 0, 1, 2, 3, 4. Distractor routes: −1, 0, 1, 2, 3 drops both endpoints, treating ≤ as if it were the strict inequality <. −2, −1, 0, 1, 2, 3, 4, 5 comes from mis-factorising as (x − 5)(x + 2), giving an upper root of 5 instead of 4. −3, −2, −1, 0, 1, 2, 3 comes from mis-factorising as (x − 4)(x + 3), giving a lower root of −3 instead of −2.
- (c) {x : x ≤ −4} ∪ {x : x ≥ 4} — Rearrange so one side is zero: x² − 16 ≥ 0, then factorise: (x − 4)(x + 4) ≥ 0. The critical values are x = −4 and x = 4. Since the coefficient of x² is positive, the graph is a U-shape that is on or above the x-axis outside its roots, so the solution is x ≤ −4 or x ≥ 4, written as {x : x ≤ −4} ∪ {x : x ≥ 4}. Distractor routes: {x : −4 ≤ x ≤ 4} takes the region BETWEEN the roots, which is where x² − 16 is negative, the opposite region. {x : x ≥ 4} keeps only the positive square root and drops the negative branch entirely. {x : x ≤ 4} comes from a sign error, treating the inequality as if it were x² ≤ 16.
- (c) x ≥ 3 — Method: collect the number terms first; the x term is negative, so the final step multiplies both sides by −1, and that is the one step that turns the inequality sign round. Working: subtracting 5 from both sides of 5 − x ≤ 2 gives −x ≤ −3; multiplying both sides by −1 turns −x into x and −3 into 3, and because the multiplier is negative the ≤ becomes ≥, so x ≥ 3. Answer: x ≥ 3. The distractors: x ≤ 3 comes from multiplying by −1 without turning the sign round, the commonest slip on this type; x ≤ −3 comes from reading −x ≤ −3 as though the minus sign could simply be rubbed off the left-hand side; x ≥ −3 comes from turning the sign round correctly but leaving the right-hand side at −3 instead of multiplying it by −1 as well.
- (d) x ≤ 30 — Method: add the fixed cost to the cost of x balloons, set that total against the £150 limit with the sign that 'at most' calls for, then solve. Working: the total spend is 60 + 3x pounds, so 60 + 3x ≤ 150; subtracting 60 from both sides gives 3x ≤ 90; dividing both sides by 3, a positive number, gives x ≤ 30. Answer: x ≤ 30. The distractors: x ≤ 90 comes from taking the cake off the budget and stopping at 3x ≤ 90, reading the 90 as a number of balloons when it is the money left for them; x ≤ 50 comes from dividing the whole £150 by 3 and leaving the cake out of the calculation altogether; x ≥ 30 comes from reading 'at most' as 'at least', which reverses the condition.
- (b) x ≤ 5 — Method: divide out the bracket first, then undo the number term; the inequality sign turns round only if both sides are multiplied or divided by a negative number. Working: dividing both sides of 2(x − 1) ≤ 8 by 2 gives x − 1 ≤ 4, and 2 is positive so the ≤ is unchanged; adding 1 to both sides gives x ≤ 5. Answer: x ≤ 5. The distractors: x ≤ 3 comes from subtracting 1 from 4 instead of adding 1 to both sides; x ≤ 4 comes from stopping at 8 ÷ 2 = 4 and never undoing the −1 inside the bracket; x < 5 comes from reading ≤ as a strict inequality, which wrongly leaves the boundary value out of the solution set.
- (b) y ≥ 2, y ≤ x, x ≤ 6 — "On or above the line y = 2" means y ≥ 2. "On or below the line y = x" means y ≤ x. "On or to the left of the line x = 6" means x ≤ 6. Together these give y ≥ 2, y ≤ x, x ≤ 6. Distractor routes: y ≤ 2, y ≤ x, x ≤ 6 flips the first inequality, describing "on or below" y = 2 instead of "on or above". y ≥ 2, y ≥ x, x ≤ 6 flips the second, describing "on or above" y = x instead of "on or below". y ≥ 2, y ≤ x, x ≥ 6 flips the third, describing "on or to the right of" x = 6 instead of "on or to the left".
- (c) 2 < x < 3 — Factorise x² − 5x + 6 = (x − 2)(x − 3), giving roots x = 2 and x = 3. Since the coefficient of x² is positive, the graph is a U-shape that dips below the x-axis between its roots. So x² − 5x + 6 < 0 for 2 < x < 3. Distractor routes: x < 2 or x > 3 takes the region OUTSIDE the roots, where the graph is above the x-axis, the opposite of what is wanted. 2 ≤ x ≤ 3 uses ≤ instead of the strict < the question asks for, wrongly including the roots themselves, where the expression equals zero, not less than zero. −3 < x < −2 comes from factorising as (x + 2)(x + 3), reversing the sign of both roots.
- (c) x ≤ 4 — Subtract 9 from both sides: −2x ≥ 1 − 9 = −8. Divide both sides by −2, flipping the inequality: x ≤ 4. A candidate who divides by −2 but forgets to flip the inequality gets x ≥ 4. A candidate who divides −8 by −2 but keeps a negative sign gets x ≤ −4. A candidate who miscalculates 1 − 9 as −10 gets x ≤ 5.
- (d) 6.0 — Subtract 1.8 from both sides: 6.3x ≤ 38.2. Divide both sides by 6.3: x ≤ 6.0634… . The question asks for the greatest value to 1 decimal place that still satisfies the inequality. Testing 6.1: 6.3 × 6.1 + 1.8 = 40.23, which is more than 40, so 6.1 fails. Testing 6.0: 6.3 × 6.0 + 1.8 = 39.6, which is no more than 40, so 6.0 works and is the greatest such value. A candidate who simply rounds 6.0634… to 1 decimal place answers 6.1, without checking that it satisfies the inequality. A candidate who adds 1.8 instead of subtracting works out 41.8 ÷ 6.3 and answers 6.6. A candidate who forgets the 1.8 altogether divides 40 by 6.3 and answers 6.3.
- (b) 2 < x < 6 — Factorise: x² − 8x + 12 = (x − 2)(x − 6), giving roots at x = 2 and x = 6. Since the coefficient of x² is positive, the graph is a U-shape that is negative, below the axis, between its roots. So P < 0 for 2 < x < 6. Distractor routes: x < 2 or x > 6 takes the region outside the roots, which is where P is positive, a profit, the opposite of a loss. 2 ≤ x ≤ 6 includes the endpoints, where P = 0 exactly, break-even rather than a loss, since the inequality is strict. −6 < x < −2 comes from factorising as (x + 2)(x + 6), reversing the sign of both roots.
- (a) x > 4 — Subtract 2x from both sides: 3x − 3 > 9. Add 3 to both sides: 3x > 12. Divide both sides by 3: x > 4. A candidate who subtracts 3 from 9 instead of adding gets 3x > 6, so x > 2. A candidate who divides correctly but wrongly flips the inequality (as if dividing by a negative) gets x < 4. A candidate who multiplies by 3 instead of dividing gets x > 36.
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