Question 1 of 24
Write x^2 - 10x + 7 in completed-square form.
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Halve -10, square the result, and compensate for the added 25.
Cambridge International AS Level · Mathematics 9709
This original practice pack develops fluent AS-level methods for completing the square, interpreting the discriminant, solving quadratic equations and inequalities, solving linear-quadratic simultaneous equations by substitution, and recognising disguised quadratics. Domain restrictions, rejected candidates, exact values and stated rounding are made explicit. It covers Cambridge 9709 Pure Mathematics 1 section 1.1 only; it is not complete AS Mathematics coverage.
2026 / 2027 · Academic review not recorded · Published 2026-09-08
AI-assisted practice — not independently academically reviewed. Answers may contain errors; check important results against your course materials.
Guided tutorial
19 lesson sections
Explanation
A quadratic polynomial has the form ax^2 + bx + c with a non-zero. The same expression can be written in expanded form, factorised form or completed-square form. Each form reveals different information: expanded form exposes the coefficients, factorised form exposes roots, and completed-square form exposes the vertex and range.
Before calculating, decide what the question asks you to reveal. Factorisation is efficient when integer or simple rational factors exist; completing the square is especially useful for a vertex, a range or an exact equation; the quadratic formula works for every quadratic equation. A final answer should respect any stated domain, interval, physical restriction or rounding instruction.
This pack is limited to Pure Mathematics 1 section 1.1. It uses elementary algebra and, in one disguised-equation discussion, familiar function notation, but it does not claim to cover the rest of Paper 1 or a complete Cambridge International AS Level route.
Explanation
For x^2 + bx, half the coefficient of x and square it: x^2 + bx = (x + b/2)^2 - (b/2)^2. Hence x^2 + bx + c = (x + b/2)^2 + c - b^2/4. Expanding the completed square is a reliable check.
When a is not 1, first factor a from both terms containing x: ax^2 + bx + c = a[x^2 + (b/a)x] + c. Complete the square inside the bracket and remember that the correction term is multiplied by a. Equivalently, ax^2 + bx + c = a(x + b/(2a))^2 + c - b^2/(4a).
Because a square is non-negative for real x, the completed-square form also gives an extremum. If a is positive, the vertex is a minimum; if a is negative, it is a maximum.
Worked example
Write 3x^2 - 12x + 7 in completed-square form. Factor 3 from the x-terms: 3(x^2 - 4x) + 7. Inside the bracket, x^2 - 4x = (x - 2)^2 - 4.
Therefore 3x^2 - 12x + 7 = 3[(x - 2)^2 - 4] + 7 = 3(x - 2)^2 - 5. Expanding gives 3(x^2 - 4x + 4) - 5 = 3x^2 - 12x + 7, so the rewrite is verified.
The minimum value is -5, attained when x = 2. Thus the vertex of y = 3x^2 - 12x + 7 is (2, -5), its axis of symmetry is x = 2, and its range for real x is y >= -5.
Explanation
In y = a(x - h)^2 + k, the vertex is (h, k) and the axis of symmetry is x = h. Notice the sign: (x + 4)^2 means h = -4. The sign of a tells whether the graph opens upward or downward, while |a| affects its steepness.
A sketch can be planned without plotting many points. Mark the vertex, the axis, the y-intercept found by setting x = 0, and any real x-intercepts found by setting y = 0. If there are no real x-intercepts, the entire graph remains on one side of the x-axis.
When the domain is restricted, the range may no longer start or end at the unrestricted vertex. Check whether the vertex's x-coordinate belongs to the stated domain; if it does not, compare the values permitted at the domain boundary or endpoints.
Worked example
For f(x) = -2x^2 - 8x + 3, complete the square: f(x) = -2(x^2 + 4x) + 3 = -2[(x + 2)^2 - 4] + 3 = -2(x + 2)^2 + 11.
On the unrestricted real domain, the vertex is (-2, 11), the axis is x = -2, and 11 is the maximum because the coefficient of the square is negative. The y-intercept is 3.
If the domain is x >= 0, the vertex is excluded. For x >= 0 the function decreases from f(0) = 3 and has no lower bound, so its restricted range is y <= 3. This illustrates why a range statement must include its domain.
Explanation
For ax^2 + bx + c = 0, the discriminant is D = b^2 - 4ac. It is the expression beneath the square root in the quadratic formula. If D > 0 there are two distinct real roots; if D = 0 there is one repeated real root; if D < 0 there are no real roots.
The discriminant also describes contact between y = ax^2 + bx + c and the x-axis. Two roots mean two crossings, a repeated root means the graph touches the axis at its vertex, and no real roots mean no contact. This conclusion concerns real roots only.
In parameter problems, substitute the coefficients carefully and solve an inequality or equation in the parameter. Pay attention when dividing an inequality by a negative number, because its direction reverses.
Worked example
Find the values of k for which x^2 + (k - 1)x + k = 0 has two distinct real roots. Here a = 1, b = k - 1 and c = k, so D = (k - 1)^2 - 4k.
Simplifying gives D = k^2 - 6k + 1. Two distinct real roots require D > 0. The boundary equation k^2 - 6k + 1 = 0 has roots k = 3 - 2sqrt(2) and k = 3 + 2sqrt(2).
Since the coefficient of k^2 is positive, k^2 - 6k + 1 is positive outside its two roots. Therefore k < 3 - 2sqrt(2) or k > 3 + 2sqrt(2). Equality is excluded because it would give a repeated root rather than two distinct roots.
Explanation
First rearrange to ax^2 + bx + c = 0. Factorise when a product of linear factors is visible, then use the zero-product rule. Completing the square is useful when the equation is close to a perfect square or when its structure matters. The formula x = [-b +/- sqrt(b^2 - 4ac)]/(2a) is universal when a is non-zero.
Keep exact values such as (5 + sqrt(13))/2 unless a decimal is requested. If rounding is requested, retain extra calculator digits until the final line and state both roots to the required accuracy. A decimal instruction such as '3 significant figures' is part of the answer, not an optional presentation choice.
Always check whether algebra has introduced candidates that the original problem forbids. Pure polynomial manipulation does not normally create extraneous roots, but squaring, clearing a variable denominator or reversing a substitution can do so.
Worked example
Solve 2x^2 - 8x - 3 = 0 exactly. Divide by 2 to obtain x^2 - 4x - 3/2 = 0, then write x^2 - 4x as (x - 2)^2 - 4.
The equation becomes (x - 2)^2 - 4 - 3/2 = 0, so (x - 2)^2 = 11/2. Taking both square-root branches gives x - 2 = +/-sqrt(11/2).
Thus x = 2 +/- sqrt(22)/2. Substitution into 2x^2 - 8x - 3 verifies both exact roots; neither branch may be discarded because the original equation imposes no domain restriction.
Worked example
Solve 5x^2 + 3x - 7 = 0, giving each root to 3 significant figures. With a = 5, b = 3 and c = -7, the discriminant is 3^2 - 4(5)(-7) = 149.
The exact roots are x = (-3 +/- sqrt(149))/10. Their calculator values are approximately 0.92065556 and -1.52065556.
Only now round to 3 significant figures: x = 0.921 or x = -1.52. Rounding sqrt(149) too early can change a final digit, so intermediate values should be retained.
Explanation
To solve a quadratic inequality, move every term to one side and find the critical values where the quadratic equals zero. These roots divide the number line into intervals on which the sign cannot change. Determine the sign in each interval from the graph's orientation or by testing one value.
For a positive leading coefficient and two real roots r1 < r2, the quadratic is positive outside the roots and negative between them. For a negative leading coefficient the signs reverse. Include a root for <= or >=, but exclude it for < or >.
Do not manipulate a quadratic inequality as if it were a linear equation by taking square roots without considering signs. A sign chart or parabola argument keeps disconnected solution intervals and endpoints visible.
Worked example
Solve 2x^2 - 5x - 3 <= 0. Factorise: 2x^2 - 5x - 3 = (2x + 1)(x - 3), so the boundary values are x = -1/2 and x = 3.
The leading coefficient is positive, so the graph opens upward. The expression is non-positive between the roots. For example, at x = 0 its value is -3, confirming the required sign there.
The solution is -1/2 <= x <= 3. Both endpoints are included because the original relation permits equality. If the relation had been < 0, the same interval would have open endpoints.
Explanation
When one equation is linear and the other quadratic, rearrange the linear equation to express one variable in terms of the other. Substitute into the quadratic equation, solve the resulting quadratic, and then recover the corresponding second coordinate for each root.
A quadratic may give zero, one or two real intersection points. Keep the pairing correct: each x-value must be matched with the y-value obtained from the same substitution. Check every ordered pair in both original equations.
If the algebra produces no real roots, the two real graphs do not intersect. A repeated root represents one point of contact. These interpretations require no diagram, and every question in this pack supplies all needed equations in text.
Worked example
Solve y = x + 1 and y = x^2 - 3x + 1 simultaneously. Equating the two expressions for y gives x + 1 = x^2 - 3x + 1, hence x^2 - 4x = 0.
Factorising gives x(x - 4) = 0, so x = 0 or x = 4. Using y = x + 1 gives y = 1 or y = 5 respectively.
The solutions are the ordered pairs (0, 1) and (4, 5). Checking in the quadratic equation gives 1 = 0 - 0 + 1 and 5 = 16 - 12 + 1, so both pairs are valid.
Worked example
Solve x + y = 1 and x^2 + y^2 = 25 simultaneously. From the linear equation, y = 1 - x. Substitute this into the second equation.
Then x^2 + (1 - x)^2 = 25, so 2x^2 - 2x - 24 = 0. Dividing by 2 gives x^2 - x - 12 = 0 = (x - 4)(x + 3). Thus x = 4 or x = -3.
Using y = 1 - x gives the ordered pairs (4, -3) and (-3, 4). Each satisfies both original equations: its coordinates sum to 1 and its squared coordinates sum to 25.
Explanation
An equation is disguised quadratic when the same expression appears squared and to the first power. For example, x^4 - 5x^2 + 4 = 0 is quadratic in x^2. Set u = x^2, solve u^2 - 5u + 4 = 0, and then reverse the substitution.
The reverse step has its own domain. If u = x^2 for real x, then u must be non-negative and each positive u gives two x-values. If u = sqrt(x), then u >= 0 and x = u^2. If u is a reciprocal, excluded denominator values from the original equation remain excluded.
After reversing a substitution, check every candidate in the original equation. This is essential when squaring was used, because squaring can turn two unequal expressions with opposite signs into equal squares and create an extraneous root.
Worked example
Solve x^4 - 13x^2 + 36 = 0 for real x. Let u = x^2, so u >= 0. The equation becomes u^2 - 13u + 36 = 0.
Factorising gives (u - 4)(u - 9) = 0, so u = 4 or u = 9. Both values satisfy the required domain u >= 0.
Reverse the substitution: x^2 = 4 gives x = +/-2, and x^2 = 9 gives x = +/-3. Therefore the four real solutions are x = -3, -2, 2 or 3, and direct substitution verifies each one.
Worked example
Solve x - 5sqrt(x) + 6 = 0 for real x. The square root requires x >= 0. Let u = sqrt(x), so u >= 0 and x = u^2. The equation becomes u^2 - 5u + 6 = 0.
Factorising gives (u - 2)(u - 3) = 0, so u = 2 or u = 3. Both respect u >= 0, producing x = 4 or x = 9. Substitution in the original equation confirms both.
Contrast this with solving sqrt(x + 2) = x by squaring. The domain requires x >= 0, and squaring gives x^2 - x - 2 = 0, with candidates 2 and -1. The candidate -1 is extraneous because it violates the original domain; only x = 2 remains after checking.
Worked example
The line y = 2x + k meets the parabola y = x^2 + 1 at exactly one point. Equating gives x^2 - 2x + (1 - k) = 0. Exactly one point means a repeated real root, so its discriminant must be zero.
Here D = (-2)^2 - 4(1)(1 - k) = 4k. Therefore k = 0. The repeated equation is x^2 - 2x + 1 = (x - 1)^2 = 0, so x = 1 and y = 2.
The unique point is (1, 2). It satisfies y = 2x and y = x^2 + 1. This example connects simultaneous equations, the discriminant and repeated-root geometry without requiring a diagram.
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Question 1 of 24
Halve -10, square the result, and compensate for the added 25.
Quick recall
Card 1 of 14
Front
Review the essentials
Completed square: factor out a when needed, halve the inner x-coefficient, and verify by expansion. Read the vertex from a(x - h)^2 + k as (h, k), then use the stated domain before declaring a range.
Discriminant and equations: compute b^2 - 4ac with brackets around negative coefficients. Distinguish D > 0, D = 0 and D < 0. Choose an efficient method, preserve exact forms unless decimals are requested, and round only on the final line.
Inequalities and simultaneous equations: put the quadratic on one side, find boundary roots, determine interval signs and treat endpoints correctly. In a linear-quadratic pair, substitute, recover every matching coordinate and check each ordered pair in both equations.
Disguised quadratics: define the temporary variable clearly, record its domain, solve the quadratic, reverse the substitution and reject invalid or extraneous candidates. Excluded scope: this pack does not cover the other Pure Mathematics 1 sections, other components, or complete AS Mathematics.
Authorship: original ai assisted.
Original practice, not an official examination paper. Readnary is not affiliated with the awarding body. Prepared with AI assistance.