The Quadratic Formula: Derivation by Completing the Square

Most students memorise the quadratic formula (x = [−b ± √(b²−4ac)] / 2a) by rote. The formula appears in the appendix of every algebra textbook; it is printed at the top of many exam papers. Students who have derived it know not just the formula but the technique — completing the square — which is useful far beyond this single result.

The starting point

ax² + bx + c = 0, where a ≠ 0.

Step 1: divide through by a.

x² + (b/a)x + c/a = 0

Step 2: move the constant to the right side.

x² + (b/a)x = −c/a

Step 3: complete the square on the left side. We add (b/2a)² to both sides.

x² + (b/a)x + (b/2a)² = −c/a + (b/2a)²

The left side is now a perfect square trinomial:

(x + b/2a)² = b²/4a² − c/a

Step 4: write the right side over a common denominator.

(x + b/2a)² = b²/4a² − 4ac/4a² = (b² − 4ac)/4a²

Step 5: take the square root of both sides.

x + b/2a = ± √[(b² − 4ac)/4a²] = ± √(b² − 4ac) / 2a

Step 6: solve for x.

x = −b/2a ± √(b² − 4ac) / 2a = [−b ± √(b² − 4ac)] / 2a

The discriminant

The quantity b² − 4ac under the radical is the discriminant. Its sign determines the nature of the roots:

b² − 4ac > 0: two distinct real roots. b² − 4ac = 0: one real root (a "repeated" root, where both roots are equal to −b/2a). b² − 4ac < 0: no real roots (two complex conjugate roots — real part −b/2a, imaginary part ±√(4ac − b²)/2a).

Why completing the square

Completing the square — converting x² + px + q to (x + p/2)² + (q − p²/4) — works for any quadratic and any coefficients. The formula is its application to the general case. Students who understand the technique can solve quadratics by completing the square directly, without the formula, which is sometimes faster and always available when the formula is forgotten.