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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.

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Linear and quadratic inequalities worksheet — GCSE Higher

Maths · GCSE Higher · A22 · Linear and quadratic inequalities · 15 questions

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1.Solve the inequality x² ≥ 16, giving your answer using set notation.
(a){x : x ≤ 4}
(b){x : x ≥ 4}
(c){x : x ≤ −4} ∪ {x : x ≥ 4}
(d){x : −4 ≤ x ≤ 4}
2.Solve the inequality 5 − x ≤ 2.
(a)x ≤ −3
(b)x ≥ −3
(c)x ≥ 3
(d)x ≤ 3
3.Solve the inequality 3x − 1 ≤ 11.
(a)x ≥ 4
(b)x ≤ 4
(c)x ≤ 12
(d)x ≤ 10/3
4.Harry will spend at most £150 on a party. The cake costs £60 and each helium balloon costs £3. Solve an inequality to find all the possible numbers of balloons, x, that he can buy.
(a)x ≥ 30
(b)x ≤ 50
(c)x ≤ 90
(d)x ≤ 30
5.Solve the inequality x² + 6x + 9 > 0.
(a)x < −3 only
(b)every value of x, with no exceptions
(c)every value of x except x = −3
(d)x > −3 only
6.Solve the inequality 2(x − 1) ≤ 8.
(a)x ≤ 4
(b)x ≤ 5
(c)x < 5
(d)x ≤ 3
7.A candidate solves the inequality x² > 9 and writes: "x² > 9, so x > 3." Which of these is a correct comment on the candidate's work?
(a)Wrong — the correct solution is −3 < x < 3.
(b)Right, but x = −3 and x = 3 should also work.
(c)Right — square rooting both sides gives x > 3.
(d)Wrong — the correct solution is x < −3 or x > 3.
8.Solve the inequality 2(3x − 1) ≥ 4x + 8.
(a)x ≥ 4.5
(b)x ≥ 5
(c)x ≥ 1
(d)x ≥ 20
9.Solve the inequality 4x + 1 > 2x + 9.
(a)x > 5
(b)x < 5
(c)x < 4
(d)x > 4
10.The solution set of an inequality is described as: x is less than −3, or x is greater than or equal to 1. Write this solution set using set notation.
(a){x : x < −3} ∪ {x : x ≥ 1}
(b){x : −3 < x ≤ 1}
(c){x : x > −3} ∪ {x : x ≤ 1}
(d){x : x ≤ −3} ∪ {x : x > 1}
11.Work out the integer values of x that satisfy x² − 2x − 8 ≤ 0.
(a)−1, 0, 1, 2, 3
(b)−2, −1, 0, 1, 2, 3, 4
(c)−2, −1, 0, 1, 2, 3, 4, 5
(d)−3, −2, −1, 0, 1, 2, 3
12.A region R satisfies y ≤ x + 1, x + y ≤ 5 and y ≥ 0. Which of these correctly describes R in words?
(a)Below y = x + 1, below x + y = 5, below y = 0
(b)Below y = x + 1, above x + y = 5, above y = 0
(c)Above y = x + 1, below x + y = 5, above y = 0
(d)Below y = x + 1, below x + y = 5, above y = 0
13.Solve the inequality 9 − 2x ≥ 1.
(a)x ≤ 5
(b)x ≥ 4
(c)x ≤ 4
(d)x ≤ −4
14.Solve the inequality 5x − 3 > 2x + 9.
(a)x > 4
(b)x > 2
(c)x < 4
(d)x > 36
15.A company's weekly profit, P thousand pounds, when it makes x thousand items is modelled by P = x² − 8x + 12, for x ≥ 0. Work out the values of x for which the company makes a loss.
(a)x < 2 or x > 6
(b)2 < x < 6
(c)−6 < x < −2
(d)2 ≤ x ≤ 6

Answer key

(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.
(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.
(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.
(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 ≤ 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.
(d) Wrong — the correct solution is x < −3 or x > 3. — Squaring or unsquaring an inequality is not a safe one-step move: the solution must be split into two branches, since a number less than −3 also squares to more than 9. So x > 3 finds only half the solution set. The full answer is x < −3 or x > 3. Distractor routes: "Right — square rooting both sides gives x > 3" treats the square root of an inequality the same as the square root of an equation and misses the negative branch entirely. "Wrong — the correct solution is −3 < x < 3" applies the between-the-roots pattern that belongs to the opposite inequality, x² < 9. "Right, but x = −3 and x = 3 should also work" treats the inequality as if it also allowed equality, when x² > 9 is strict and excludes both x = 3 and x = −3.
(b) x ≥ 5 — Expand the bracket: 2(3x − 1) = 6x − 2, so the inequality is 6x − 2 ≥ 4x + 8. Subtract 4x from both sides and add 2 to both sides: 2x ≥ 10. Divide both sides by 2: x ≥ 5. A candidate who only multiplies the 3x by 2 and forgets to multiply the −1 gets 6x − 1 ≥ 4x + 8, leading to x ≥ 4.5. A candidate who adds 4x instead of subtracting it gets 10x ≥ 10, leading to x ≥ 1. A candidate who multiplies by 2 instead of dividing gets x ≥ 20.
(d) x > 4 — Method: collect the x terms on one side and the numbers on the other, then divide by the coefficient of x; dividing by a positive number leaves the sign as it is. Working: subtracting 2x from both sides of 4x + 1 > 2x + 9 gives 2x + 1 > 9; subtracting 1 from both sides gives 2x > 8; dividing both sides by 2 gives x > 4. Answer: x > 4. The distractors: x < 4 comes from turning the sign round while dividing by 2; x > 5 comes from adding the 1 to the 9 instead of subtracting it, giving 2x > 10; x < 5 comes from making both of those mistakes together.
(a) {x : x < −3} ∪ {x : x ≥ 1} — "Less than −3" stays strict, since the wording never says "or equal to": x < −3. "Greater than or equal to 1" is inclusive: x ≥ 1. These are two separate, non-overlapping ranges joined with "or", so in set notation they are combined with the union symbol: {x : x < −3} ∪ {x : x ≥ 1}. Distractor routes: {x : x ≤ −3} ∪ {x : x > 1} swaps the strict and inclusive signs, marking −3 as included and 1 as excluded, the opposite of the wording. {x : −3 < x ≤ 1} treats "or" as "and", joining the two conditions into one continuous interval between the values instead of a union of two separate ranges. {x : x > −3} ∪ {x : x ≤ 1} reverses both inequality directions; the two reversed ranges then overlap and between them cover every number on the number line, so that set is the whole of the real line rather than the two separate ranges the description asks for.
(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.
(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.
(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.
(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.
(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.

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