Printable · GCSE Higher · ages 14-16
Algebra worksheet — GCSE Higher
Fifteen questions across the algebra statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Algebra worksheet — GCSE Higher
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- 1.A rule turns each input x into an output y. An input of 1 gives an output of 1, an input of 2 gives an output of 4 and an input of 3 gives an output of 9. Work out the rule.
- 2.A rectangular photo has length (x + 3) cm and width x cm. Its area is 40 cm². Work out the value of x.
- 3.Work out the gradient of the straight line with equation y = (2x/5) − 3.
- 4.A table of values is being drawn for the graph of y = x³. Work out the value of y when x = −2.y = x
- 5.Solve the inequality x² + 6x + 9 > 0.
- 6.Expand and simplify (x − 3)²
- 7.Meera writes the statement 3(x + 4) = 3x + 12. Which of these correctly describes what she has written, with a reason?
- 8.For the graph of y = 1/x, where x cannot be zero, which of these statements is correct?
- 9.A rule multiplies the input by a fixed number and then adds a fixed number. An input of 1 gives an output of 5, and an input of 3 gives an output of 11. Work out the rule, writing the input as x and the output as y.
- 10.The curve y = x² − 3x. Use the chord between x = 1 and x = 4 to estimate the gradient of the curve at x = 2.5.y = x² − 3x
- 11.Factorise 3x² + 10x − 8.
- 12.To rearrange y = 2x − 5 to make x the subject, Sam writes: 'Add 5 to both sides to get y + 5 = 2x, then divide both sides by 2 to get x = (y + 5)/2.' Which statement about Sam's method is correct?y = 2x − 5
- 13.By completing the square, find the turning point of the curve y = 3x² + 12x + 7.y = 3x² + 12x + 7
- 14.A quadratic graph has equation y = (x − 4)². A student says this graph crosses the x-axis at two different points. Explain why the student is wrong.
- 15.The first four terms of a sequence are 4, 9, 14, 19. Work out an expression, in terms of n, for the nth term.
Answer key
- (c) y = x² — Method: test a candidate rule against every pair given, not just one — a rule that fits one pair and fails another is not the rule. Working: the outputs 1, 4, 9 rise by 3 and then by 5, so they are not going up in equal steps and the input is not simply multiplied by a fixed number; comparing each output with its own input gives 1 × 1 = 1, 2 × 2 = 4 and 3 × 3 = 9, and all three pairs fit. Answer: y = x². The distractors: y = 3x comes from fitting only the last pair, where 3 × 3 = 9, and reading that 3 as a multiplier; y = 3x − 2 comes from assuming a multiply-then-add rule and using the first step in the outputs, 4 − 1 = 3, as the multiplier — it fits the first two pairs and fails the third; y = 2x comes from fitting only the pair 2 and 4 and reading every output as double its input.
- (b) x = 5 — Area = length × width, so x(x + 3) = 40, which rearranges to x² + 3x − 40 = 0. This factorises as (x + 8)(x − 5) = 0: the two numbers in the brackets must multiply to −40 and add to +3, and the pair 8 and −5 does both. This gives x = −8 or x = 5. Since x is a length, it cannot be negative, so x = 5. A candidate who gives both solutions without rejecting the negative one, which cannot be a length, answers x = 5 or x = −8. A candidate who picks the wrong factor pair of 40, such as 10 and −4 instead of 8 and −5, gets (x + 10)(x − 4) = 0 and answers x = 4. A candidate who rejects the wrong root, keeping the negative solution instead of the positive one, answers x = −8.
- (a) 2/5 — Method: compare the equation with y = mx + c, where m is the gradient. Working: in y = (2x/5) − 3, the coefficient of x is 2/5. Answer: the gradient is 2/5. −3 comes from confusing the gradient with the y-intercept. 5/2 comes from inverting the fraction that multiplies x. −2/5 comes from wrongly carrying the negative sign from the −3 term onto the coefficient of x.
- (c) −8 — (−2)³ = (−2) × (−2) × (−2) = −8, since multiplying three negative numbers gives a negative result. A candidate who forgets the sign of a negative number when cubing it might treat (−2)³ as if it were 2³ = 8. A candidate who multiplies −2 by 3 instead of cubing it might get −2 × 3 = −6. A candidate who combines both mistakes — multiplying by 3 and dropping the sign — might get 2 × 3 = 6.
- (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.
- (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).
- (a) An identity, true for every value of x — Expanding the bracket on the left gives 3x + 12, which matches the right-hand side exactly, so the statement is true for every value of x — this makes it an identity. A candidate who reasons that any statement with an equals sign must be an equation picks that option, missing that an equation is only true for particular value(s) of x, not all of them. A candidate who confuses an identity with a formula, because both relate two expressions, picks the formula option — but a formula connects two different quantities, such as area and side length, not two equivalent forms of the same expression. A candidate who assumes it can be solved for a single value of x, as with a normal equation, picks that option, not realising there is no single solution here.
- (d) The graph never crosses either axis — Since x ≠ 0, there is no point on the graph where x = 0, so it cannot cross the y-axis; likewise 1/x is never equal to 0 for any x, so it cannot cross the x-axis either — the graph never touches either axis. A candidate who forgets the restriction x ≠ 0 might think the graph behaves like other graphs and passes through the origin, (0, 0). A candidate who correctly rules out the x-axis but forgets that x = 0 is also excluded might say the graph crosses the y-axis but never the x-axis. A candidate who only pictures the branch where x and y are both positive might say the graph has only one branch, in quadrant 1, forgetting the second branch where x and y are both negative.
- (c) y = 3x + 2 — Method: divide the change in the outputs by the change in the inputs to find the multiplier, then put one pair of values into the rule to find the number added on. Working: the output rises by 11 − 5 = 6 while the input rises by 3 − 1 = 2, so the multiplier is 6 ÷ 2 = 3; with an input of 1, 3 × 1 = 3 and the output is 5, so 2 is added. Answer: y = 3x + 2, checked against the second pair by 3 × 3 + 2 = 11. The distractors: y = 3x − 2 comes from finding the multiplier 3 and then subtracting the 2 instead of adding it; y = 2x + 3 comes from swapping the multiplier and the number added on; y = x + 4 comes from assuming the input is multiplied by 1 and using 5 − 1 = 4 as the number added on.
- (c) 2 — At x = 1, y = 1² − 3(1) = 1 − 3 = −2. At x = 4, y = 4² − 3(4) = 16 − 12 = 4. Gradient of the chord = change in y ÷ change in x = (4 − (−2)) ÷ (4 − 1) = 6 ÷ 3 = 2. Writing down the change in y, 6, and stopping there without dividing by the change in x gives 6. Losing the negative sign on y = −2 at x = 1 and treating it as +2 gives (4 − 2) ÷ (4 − 1) = 2 ÷ 3 = 2/3. Dividing the wrong way round, change in x ÷ change in y, gives (4 − 1) ÷ (4 − (−2)) = 3 ÷ 6 = 1/2.
- (d) (3x − 2)(x + 4) — For 3x² + 10x − 8, find two numbers multiplying to 3 × (−8) = −24 and adding to 10: these are 12 and −2. Rewrite: 3x² + 12x − 2x − 8 = 3x(x + 4) − 2(x + 4) = (3x − 2)(x + 4). Choosing (3x + 2)(x − 4) expands to 3x² − 10x − 8 — the correct factor pair but the wrong signs, giving the middle term the wrong sign. Choosing (x − 2)(3x + 4) expands to 3x² − 2x − 8 — the 3 is attached to the wrong bracket, changing which terms combine for the x-coefficient. Choosing (3x − 4)(x + 2) expands to 3x² + 2x − 8 — this uses 4 and 2 instead of the correct pair 12 and 2, so the middle term does not come to 10x.
- (c) Yes; x = (y + 5)/2 is right — Sam's method is correct throughout: adding 5 to both sides gives y + 5 = 2x, and dividing both sides by 2 gives x = (y + 5)/2, so Sam is right. The option giving x = (y − 5)/2 assumes 5 should be subtracted again, but 2x − 5 = y means 5 has already been subtracted, so it must be added back, not taken away a second time. The option giving x = y/2 + 5 divides only the y term by 2 and leaves the 5 unhalved, which is not a valid rearrangement. The option agreeing Sam is correct but changing step 2 to x = 2(y + 5) confuses '2x' with 'x divided by 2' — since x is multiplied by 2, the inverse is division, not multiplication. Sam's working, and his final formula x = (y + 5)/2, are both correct.
- (d) x = −2, y = −5 — 3x² + 12x + 7 rewrites as 3(x² + 4x) + 7, then as 3[(x + 2)² − 4] + 7, which simplifies to 3(x + 2)² − 5, since −3 × 4 + 7 = −5. Substituting x = −2: 3 × (−2)² = 12, 12 × (−2) = −24, so 12 − 24 + 7 = −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 = −41, 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 12 − 7 = 5 instead of 7 − 12 = −5 flips the sign of the constant, giving x = −2, y = 5 — wrong, since the completed square's constant must be evaluated as 7 minus 12, not 12 minus 7. 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.
- (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.
- (d) 5n − 1 — The common difference is 5 (9−4=5), so the expression starts 5n. To match the first term when n=1, 5×1+c=4, so c=−1: the nth term is 5n−1. A candidate who uses the first term itself as the constant, instead of first term minus the common difference, would write 5n+4 (giving 9, 14, 19, 24 — one term too high throughout). A candidate who omits the constant term altogether would write just 5n (giving 5, 10, 15, 20, not matching the sequence). A candidate who adds the common difference to n instead of multiplying would write n+5 (giving 6, 7, 8, 9, far too small).
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