Solving quadratic equations

A quadratic equation has the form ax² + bx + c = 0 (where a ≠ 0). There are three main methods for solving quadratics.

Method 1: Factorising

Look for two numbers that multiply to give ac and add to give b.

Simple case (a = 1): x² + 5x + 6 = 0. Find two numbers that multiply to 6 and add to 5 → 2 and 3. Factor: (x + 2)(x + 3) = 0. Solutions: x = −2 or x = −3.

Harder case (a ≠ 1): 2x² + 7x + 3 = 0. ac = 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7 → 1 and 6. Split the middle: 2x² + x + 6x + 3 = 0. Group: x(2x + 1) + 3(2x + 1) = 0 → (x + 3)(2x + 1) = 0. Solutions: x = −3 or x = −1/2.

Difference of two squares: x² − 16 = 0 → (x − 4)(x + 4) = 0 → x = ±4.

Method 2: Quadratic formula

For any quadratic ax² + bx + c = 0: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

This always works. Use it when factorising is difficult or when answers are not whole numbers.

Discriminant: Δ = b² − 4ac

• Δ > 0: two distinct real solutions.
• Δ = 0: one repeated (equal) solution.
• Δ < 0: no real solutions (Higher tier — indicates no intersection with x-axis).

Example: 3x² − 5x − 2 = 0. a = 3, b = −5, c = −2. Δ = 25 − 4(3)(−2) = 25 + 24 = 49. x = (5 ± 7)/6 → x = 2 or x = −1/3.

Method 3: Completing the square (Higher)

Rewrite x² + bx + c in the form (x + p)² + q. The vertex of the parabola is at (−p, q).

Method: x² + 6x − 7 = 0. x² + 6x = 7. (x + 3)² − 9 = 7 (half the x-coefficient, square it). (x + 3)² = 16. x + 3 = ±4 → x = 1 or x = −7.

Completing the square is also used to find the minimum/maximum of a quadratic and to write the vertex form for sketching.

CCEA context

CCEA Paper 1 tests factorising (non-calculator). Paper 2 tests the quadratic formula and completing the square. Higher tier only: discriminant and completing the square with a ≠ 1.