amazon.co.ukSolving quadratics by factorising, completing the square and the formula, the discriminant, quadratic inequalities and hidden quadratics - A-Level pure maths (Edexcel 9MA0 Topic 2.3, AQA 7357 B3) revised with worked examples and mark-scheme logic.
Every board examines quadratic solving early and often (Edexcel 9MA0 2.3; AQA 7357 B3; OCR A H640 3.01; WJEC/Eduqas and CCEA AS Pure Mathematics), and the first decision is method. Factorising is fastest when an integer factor pair exists: for a = 1, find two numbers with product c and sum b; for a is not 1, use the AC method - for 2x^2 - 5x - 3, compute ac = -6 and find two numbers with product -6 and sum -5, namely -6 and 1, so 2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = (2x + 1)(x - 3). Completing the square is required when the question asks for the vertex or exact surd roots. The quadratic formula x = (-b +/- sqrt(b^2 - 4ac))/2a always works and is the expected method when a question requests surd form or when factorising is impractical.
Method marks are awarded for correct factorisation or a correct substitution into the formula, and accuracy marks for the roots - so write the formula substitution explicitly, even when using a calculator, because the substitution line is where the method mark lives.
Solve 2x^2 - 5x - 3 = 0.
Line 1: ac = -6; two numbers with product -6 and sum -5: -6 and 1 (M1).
Line 2: 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3) (A1).
Line 3: (2x + 1)(x - 3) = 0, so x = -1/2 or x = 3 (A1).
Check roots by substitution: 2(9) - 5(3) - 3 = 0 and 2(1/4) - 5(-1/2) - 3 = 1/2 + 5/2 - 3 = 0. Both check out, so the factorisation is verified - a 20-second safety net.
Completing the square rewrites the quadratic so the structure is visible: x^2 + bx = (x + b/2)^2 - (b/2)^2, and in general ax^2 + bx + c = a[(x + b/2a)^2] + (c - b^2/4a). Three uses: (1) solving with exact surd roots; (2) reading the turning point - the minimum of a(x + p)^2 + q is at (-p, q) for a > 0; (3) deriving the quadratic formula and discriminant results from scratch.
The half-coefficient discipline: halve the coefficient of x, square it, and subtract it. Sign errors here are the most common mark loss, especially with negative b and with a leading coefficient that must be factored out first.
Solve x^2 - 4x + 1 = 0 by completing the square, giving exact answers.
Line 1: Half of -4 is -2; squared is 4. x^2 - 4x + 1 = (x - 2)^2 - 4 + 1 = (x - 2)^2 - 3 (M1 A1).
Line 2: (x - 2)^2 = 3, so x - 2 = +/- sqrt(3) (M1).
Line 3: x = 2 +/- sqrt(3), i.e. x = 2 + sqrt(3) or x = 2 - sqrt(3) (A1: exact surd form).
Write 2x^2 - 8x + 5 in the form a(x + p)^2 + q and state the minimum point.
Line 1: Factor out 2: 2(x^2 - 4x) + 5 (M1).
Line 2: Complete the square inside: 2[(x - 2)^2 - 4] + 5 = 2(x - 2)^2 - 8 + 5 = 2(x - 2)^2 - 3 (A1).
Line 3: Minimum at (2, -3), since 2(x - 2)^2 >= 0 with equality when x = 2 (A1).
The completed square form also proves the discriminant results: ax^2 + bx + c = 0 rearranges to (x + b/2a)^2 = (b^2 - 4ac)/4a^2, and the left side is a square, so a real solution exists only when b^2 - 4ac >= 0 - this derivation is itself a common exam question.
The discriminant b^2 - 4ac determines the nature of the roots without solving: greater than 0 gives two distinct real roots; equal to 0 gives one repeated root, which is exactly the condition for a line to be tangent to a curve; below 0 gives no real roots. Exam questions reverse this: given a condition on the roots, form an inequality in an unknown coefficient and solve it.
Two edge cases separate A* candidates. First, when the unknown is the coefficient of x^2, the equation may cease to be quadratic: kx^2 + 4x + k = 0 with k = 0 becomes the linear equation 4x = 0, which does have a (single) real root - so k = 0 must be excluded if the question demands a quadratic with two distinct roots. Second, tangency language: a line tangent to a curve means exactly one intersection point, hence discriminant zero after substitution.
Find the set of values of k for which kx^2 + 4x + (k - 3) = 0 has two distinct real roots.
Line 1: Two distinct real roots require b^2 - 4ac > 0: 16 - 4k(k - 3) > 0 (M1).
Line 2: 16 - 4k^2 + 12k > 0, so k^2 - 3k - 4 < 0 after dividing by -4 and flipping the inequality (M1: sign flip).
Line 3: (k - 4)(k + 1) < 0, so -1 < k < 4 (A1: strict inequalities).
Line 4: k is not 0, otherwise the equation is linear, so the final answer is -1 < k < 4 with k = 0 excluded (A1: edge case).
The equation 2x^2 + bx + 5 = 0 has roots alpha and beta such that alpha + beta = alpha beta. Find b.
Line 1: Sum of roots: alpha + beta = -b/2 (M1).
Line 2: Product of roots: alpha beta = 5/2 (M1).
Line 3: Given alpha + beta = alpha beta: -b/2 = 5/2, so b = -5 (A1).
Harder variants combine the two ideas: for which k does kx^2 + 8x + k = 0 have roots whose product equals twice their sum? Here alpha beta = c/a = 1 and alpha + beta = -8/k, so 1 = 2(-8/k), giving k = -16. Setting up the ratio equation from sum and product is the transferable A* skill.
Quadratic inequalities (Edexcel 9MA0 2.4; AQA 7357 B3; OCR A H640 2.01) are solved by sketching, never by algebra alone. The routine: (1) rearrange so the quadratic is on one side; (2) solve the corresponding equation for the critical values; (3) sketch the parabola - upward for a > 0; (4) read off the region where the inequality holds; (5) write the answer in correct notation, with strict inequalities for > and < (no equality) and inclusive brackets for >= and <=.
For x^2 - 5x + 6 > 0: critical values x = 2 and x = 3; the parabola opens upward, so it is positive outside the roots: x < 2 or x > 3. Reversing to < 0 gives 2 < x < 3. The trap: for a < 0, the regions swap - a negative quadratic is positive between its roots.
Combined inequalities intersect solution sets: find k such that k^2 - 3k - 4 < 0 and k > 0: the first gives -1 < k < 4, the second k > 0, so 0 < k < 4. Always solve the inequalities separately and intersect at the end.
Sketching from completed square form ties the topic together: y = 2(x - 2)^2 - 3 has minimum (2, -3), y-intercept 5 (set x = 0), and roots found by setting the square equal to 3: x = 2 +/- sqrt(3/2). A sketch annotated with intercepts and the turning point earns full marks on sketch questions, and the roots should be labelled with exact values.
A hidden quadratic is any equation that becomes quadratic under a substitution: u = x^2 for quartics like x^4 - 5x^2 + 4 = 0; u = sin(theta) for trigonometric equations; u = e^x for exponential equations; u = 3^x for 3^(2x) - 12(3^x) + 27 = 0. The method: substitute, solve the quadratic in u, then back-substitute and check the range of the substituted variable - this check is where marks are won and lost, because solutions in u may be impossible in the original variable.
Solve 2x^2 - 5x - 3 = 0, and hence solve 2sin^2(theta) - 5sin(theta) - 3 = 0 for 0 <= theta < 2pi.
Line 1: From the first part: 2x^2 - 5x - 3 = (2x + 1)(x - 3) = 0, so x = -1/2 or x = 3 (M1 A1).
Line 2: Substitute u = sin(theta): sin(theta) = -1/2 or sin(theta) = 3 (M1: substitution).
Line 3: sin(theta) = 3 is impossible since sin(theta) is between -1 and 1, so reject it (B1: range check).
Line 4: sin(theta) = -1/2: reference angle pi/6, solutions in quadrants 3 and 4: theta = pi + pi/6 = 7pi/6 and theta = 2pi - pi/6 = 11pi/6 (M1 A1 A1).
Exam technique across the topic: (1) write the substitution explicitly - the M1 depends on it; (2) never skip the range check; (3) for equation chains (solve, hence solve), use the first result rather than re-solving; (4) keep surd answers exact until a decimal is requested; (5) when a discriminant inequality arises, watch the sign flip when dividing by a negative and watch the k = 0 edge case. Board coverage: Edexcel 9MA0 Topic 2.3-2.4, AQA 7357 B3, OCR A H640 3.01 and 2.01, WJEC/Eduqas AS Unit 1 algebra, CCEA AS1/A21 - the techniques are identical, only the question wording differs.