Stewdent.eu
▾

Completing the Square

Completing the square rewrites a quadratic expression ax² + bx + c in the form a(x + p)² + q. This technique is used to solve quadratics, find the vertex of a parabola, and derive the quadratic formula.

Last updated: 11 November 2025

★★★★☆ Level 4mathAlgebraQuadraticcompleting-the-squarevertex4 min read

Report an error

Why Complete the Square?

The quadratic formula solves any quadratic, but completing the square also:

  • Reveals the vertex (turning point) of a parabola directly
  • Solves equations without needing to factorise
  • Helps sketch y=x2+bx+cy = x^2 + bx + c accurately

Method for x2+bx+cx^2 + bx + c (coefficient of x2x^2 is 1)

  1. Halve the coefficient of xx to get p=b/2p = b/2
  2. Write (x+p)2−p2+c(x + p)^2 - p^2 + c

Complete the square for x² + 6x + 5

b=6b = 6, so p=3p = 3

(x+3)2−9+5(x + 3)^2 - 9 + 5

=(x+3)2−4= (x + 3)^2 - 4

Check by expanding: (x+3)2−4=x2+6x+9−4=x2+6x+5(x+3)^2 - 4 = x^2 + 6x + 9 - 4 = x^2 + 6x + 5 ✓

Solving by Completing the Square

Solve x² + 6x + 5 = 0

From above: (x+3)2−4=0(x + 3)^2 - 4 = 0

(x+3)2=4(x + 3)^2 = 4

x+3=±2x + 3 = \pm 2

x=−3+2=−1x = -3 + 2 = -1 \quad or x=−3−2=−5\quad x = -3 - 2 = -5

Reading off the Vertex

The form (x+p)2+q(x + p)^2 + q tells us the minimum point of the parabola is (−p,q)(-p, q).

For (x+3)2−4(x + 3)^2 - 4: vertex is (−3,−4)(-3, -4).

Tip

The minimum value of (x+p)2+q(x + p)^2 + q is qq, because a square is always ≥ 0. The vertex is therefore the point where x=−px = -p.

When the Coefficient of x2x^2 is Not 1

Factor it out first:

2x2+8x+3=2(x2+4x)+3=2[(x+2)2−4]+3=2(x+2)2−52x^2 + 8x + 3 = 2(x^2 + 4x) + 3 = 2[(x+2)^2 - 4] + 3 = 2(x+2)^2 - 5

Real-World Examples

Real world: maximum height of a thrown ball

A ball's height (in metres) is h=−4t2+16t+2h = -4t^2 + 16t + 2.

Complete the square: h=−4(t2−4t)+2=−4[(t−2)2−4]+2=−4(t−2)2+18h = -4(t^2 - 4t) + 2 = -4[(t-2)^2 - 4] + 2 = -4(t-2)^2 + 18.

The ball reaches its maximum height of 18 m at t=2t = 2 seconds.

Real world: highest point of a bridge arch

A parabolic arch is described by y=−x2+8xy = -x^2 + 8x (metres). Completing the square:

y=−(x2−8x)=−[(x−4)2−16]=−(x−4)2+16y = -(x^2 - 8x) = -[(x-4)^2 - 16] = -(x-4)^2 + 16.

The arch peaks at 16 m above road level, directly above x=4x = 4 m. Engineers use this to ensure clearance for vehicles or river traffic.

Real world: focal point of a satellite dish

A satellite dish cross-section follows y=14x2y = \frac{1}{4}x^2. The vertex form shows the vertex at (0,0)(0, 0) and the focal point at (0,1)(0, 1) — 1 m above the dish centre.

The receiver must be placed exactly there to capture the concentrated signal. Any offset reduces reception strength significantly.

Practice Problems

Complete the square for x² − 4x + 7.

Show answer
p = −4/2 = −2. (x−2)² − 4 + 7 = (x−2)² + 3. Vertex: (2, 3).

Solve x² − 4x + 3 = 0 by completing the square.

Show answer
(x−2)² −4 + 3 = 0 → (x−2)² = 1 → x−2 = ±1 → x = 3 or x = 1.

Write x² + 10x − 3 in completed square form and state the minimum value.

Show answer
(x+5)² − 25 − 3 = (x+5)² − 28. Minimum value is −28 (when x = −5).