Lesson 4: Quadratic Equations

Quadratic equations — equations with x² — are the centrepiece of GCSE Higher algebra and the foundation of A-level. Their graphs are parabolas, their solutions come in pairs, and they model everything from projectile motion to profit maximisation. This lesson covers all three solution methods (factorising, the quadratic formula, completing the square), the discriminant, and quadratic graphs — with the fluency that examiners expect. (Quadratics are Higher tier, except for simple factorising.)

What makes an equation quadratic?

A quadratic equation has the form ax² + bx + c = 0, with a ≠ 0. The x² term is what makes it quadratic (from *quadratus*, Latin for 'square'). Unlike linear equations with one solution, quadratics can have two, one, or no real solutions — picture a parabola crossing the x-axis twice, touching it once, or missing it entirely.

Before solving, always rearrange into the standard form ax² + bx + c = 0. For x² + 5x = 14: subtract 14 → x² + 5x − 14 = 0. For 2x² = 8x: bring all terms left → 2x² − 8x = 0 (then factorise: 2x(x − 4) = 0, so x = 0 or x = 4). Never divide both sides by x — you would lose the solution x = 0. This is the single most common serious error in the topic.

Quadratics model real situations: the height of a cricket ball hit at Lord's follows h = 20t − 5t² (t in seconds); setting h = 0 gives 5t(4 − t) = 0, so t = 0 (hit) or t = 4 (lands) — the ball is in the air for 4 seconds. The two solutions have physical meaning: start and finish.

Solving by factorising

Factorising is the fastest method when it works: write the quadratic as two brackets, then use the zero-product rule — if AB = 0, then A = 0 or B = 0. For x² + 5x − 14 = 0: factorise → (x + 7)(x − 2) = 0, so x + 7 = 0 or x − 2 = 0, giving x = −7 or x = 2. Check: (−7)² + 5(−7) − 14 = 49 − 35 − 14 = 0. ✓

Worked example: solve 2x² + 7x + 3 = 0. Find factor pairs of 2 × 3 = 6 with sum 7: 6 and 1. Split the middle term: 2x² + 6x + x + 3 = 0 → 2x(x + 3) + 1(x + 3) = 0 → (2x + 1)(x + 3) = 0. So x = −1/2 or x = −3. The 'split the middle term' method works for every factorisable quadratic with a ≠ 1.

Not all quadratics factorise over the integers: x² + x + 1 = 0 has no integer factors, and x² − 5x + 3 = 0 does not factorise neatly. When factorising fails — or when the question says 'give answers to 2 decimal places' — use the formula. Recognising when to switch methods is a key exam skill.

The quadratic formula

The quadratic formula solves every quadratic: for ax² + bx + c = 0, x = (−b ± √(b² − 4ac)) / 2a. Memorise it — examiners expect you to quote it from memory. Identify a, b, c carefully, watching signs: for 2x² − 5x − 3 = 0, a = 2, b = −5, c = −3.

Worked example: solve x² − 6x + 4 = 0, giving answers to 2 decimal places. a = 1, b = −6, c = 4. Then:

x = (6 ± √(36 − 16)) / 2 = (6 ± √20) / 2 = (6 ± 4.4721) / 2
x = 10.4721 / 2 = 5.24   or   x = 1.5279 / 2 = 0.76

Keep full calculator precision until the final rounding — early rounding loses the accuracy mark. The ± gives two solutions, but they may coincide or be non-real (see the discriminant below). Always present both solutions, and check by substitution when time permits. Exam tip: write the formula, then substitute with brackets — x = (−(−6) ± √((−6)² − 4(1)(4))) / 2(1) — so sign errors are visible and method marks are safe. This is a calculator-paper method; on the non-calculator paper you will be expected to factorise or complete the square.

Completing the square

Completing the square rewrites a quadratic in the form (x + p)² + q, revealing the vertex of the parabola. For x² + 6x + 5: halve the x-coefficient (6 → 3), so (x + 3)² = x² + 6x + 9 — we have added 9, so subtract it: (x + 3)² − 9 + 5 = (x + 3)² − 4. The minimum value is −4, at x = −3 — read straight off.

Worked example: solve x² + 8x + 7 = 0 by completing the square. (x + 4)² − 16 + 7 = 0 → (x + 4)² = 9 → x + 4 = ±3 → x = −1 or x = −7. This method also proves the quadratic formula (complete the square on ax² + bx + c = 0 in general) — a beautiful piece of algebra worth seeing once.

For a ≠ 1, factor out a first: 2x² + 12x + 10 = 2(x² + 6x + 5) = 2[(x + 3)² − 4] = 2(x + 3)² − 8. Completing the square is essential at A-level (for integration, circles, and optimisation) — master it now and A-level starts easier.

The discriminant and quadratic graphs

The discriminant, D = b² − 4ac, predicts the solutions without solving: D > 0 → two distinct real roots (the parabola crosses the axis twice); D = 0 → one repeated root (it touches the axis); D < 0 → no real roots (it misses the axis entirely).

The graph of y = ax² + bx + c is a parabola: ∪-shaped if a > 0, ∩-shaped if a < 0. The y-intercept is c; the roots (where it crosses the x-axis) are the solutions of ax² + bx + c = 0; the vertex is the turning point, at x = −b/2a.

Worked example: a farmer in Devon has 40 m of fencing for a rectangular pen against a straight river (no fence needed on the river side). Maximise the area. If the sides perpendicular to the river have length x, the parallel side is 40 − 2x, so A = x(40 − 2x) = −2x² + 40x. The vertex is at x = −40/(2 × −2) = 10 m, giving A = 10 × 20 = 200 m². Quadratics optimise — this is why businesses and engineers love them.

Worked example: the product of two consecutive even numbers is 168. Find the numbers. Let them be n and n + 2: n(n + 2) = 168 → n² + 2n − 168 = 0 → (n + 14)(n − 12) = 0 → n = 12 or n = −14. Both pairs work: 12 and 14, or −14 and −12 (unless the question says 'positive'). Read the wording carefully before rejecting a solution.

Problem solving — quadratics in context, planned

Multi-step quadratic problems follow a fixed rhythm: define the variable, form the equation, solve, then interpret (including rejecting impossible roots). Worked example: a rectangular photo has length 4 cm more than its width. Its area is 96 cm². Find its dimensions. Let the width be w: w(w + 4) = 96 → w² + 4w − 96 = 0 → factorise: (w + 12)(w − 8) = 0 → w = 8 or w = −12. A width of −12 cm is impossible, so w = 8 cm and the length is 12 cm. Check: 8 × 12 = 96. ✓ The rejection sentence is part of the solution — examiners award it.

Worked example: a ball is thrown and its height after t seconds is h = 12t − 5t² metres. When is it 7 metres up? 12t − 5t² = 7 → 5t² − 12t + 7 = 0 → factorise: (5t − 7)(t − 1) = 0 → t = 1 or t = 1.4. Here both roots are meaningful: the ball passes 7 m on the way up (t = 1 s) and again on the way down (t = 1.4 s). Context decides: geometry problems usually reject a root; motion problems often keep both. Never reject by habit — reject by reasoning, and write the reason.

Common mistakes

• Dividing by x. Solving x² = 5x by dividing gives x = 5 and silently loses x = 0. Factorise instead: x(x − 5) = 0.
• Sign errors in the formula. For 2x² − 5x − 3 = 0, b = −5, so −b = +5 and b² = 25 (not −25). Substitute with brackets.
• Forgetting the ±. The formula usually gives two answers; writing only the positive root halves the marks.
• Stopping at (x + 4)² = 9. Completing the square is a method, not the answer — take the square root (with ±) and finish.
• Accepting impossible answers. A width of −12 cm cannot be right for a rectangle; reject it in the context, and say why.
• Rounding inside the formula. Keep √20 as 4.4721... in your calculator memory until the final line.

Exam Experience

Question 1. Solve x² − 2x − 15 = 0. [2 marks]

Question 2. Show that the equation 2x² + 4x + 5 = 0 has no real solutions. [3 marks]

Question 3. A rectangle has length 3 cm more than its width. The area of the rectangle is 40 cm². Work out the width of the rectangle. [3 marks]

Mark scheme.

• Q1: M1 for factorising (x − 5)(x + 3) = 0 (or a correct use of the formula). A1 for both solutions, x = 5 and x = −3.
• Q2: M1 for using the discriminant b² − 4ac with a = 2, b = 4, c = 5. M1 for evaluating: 16 − 40 = −24. A1 for the conclusion: the discriminant is negative, so there are no real solutions. (Equally acceptable: completing the square to get 2(x + 1)² + 3, which is always positive.)
• Q3: M1 for forming the equation w(w + 3) = 40, i.e. w² + 3w − 40 = 0. M1 for solving, (w + 8)(w − 5) = 0. A1 for w = 5 cm (the negative root rejected as a length).