Completing the Square

Rewriting a quadratic as a single perfect square — the halve-and-square rule, worked through equations with a leading coefficient of 1, other coefficients, and no real solution.

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Completing the Square — Moosa Academy

A quadratic equation is hard to solve because the unknown appears both squared and on its own. Completing the square removes that difficulty by rewriting the equation as a single perfect square, after which taking a square root finishes the job.

Concept The idea

(x + m)^{2} = k \quad \Longrightarrow \quad x + m = \pm\sqrt{k}

x² 3x 3x 9 the missing corner
x^{2} + 6x fills all of the square except one corner.
That corner has area 9 = (6/2)^{2} .
Add it and the figure becomes the complete square (x+3)^{2} .

The name is literal: an incomplete square is being completed. Recall the identity (x + m)^{2} = x^{2} + 2mx + m^{2} — this method runs it backwards.

Note The three steps
  1. Make the coefficient of x^{2} equal to 1, dividing the whole equation if necessary.
  2. Take the coefficient of x , halve it, square it, and add that to both sides.
  3. Write the left side as a squared bracket and take the square root of both sides, remembering \pm .

Adding to only one side is the most common mistake, and it changes the equation.

Example A leading coefficient of 1

Solve x^{2} + 6x + 5 = 0 .

The coefficient of x^{2} is already 1.
Half of 6 is 3, and 3^{2} = 9 . Add 9 to both sides:
x^{2} + 6x + 9 + 5 = 9
Move the constant across: x^{2} + 6x + 9 = 4
(x + 3)^{2} = 4
x + 3 = \pm 2
⟹ x = −1 or x = −5
Example A leading coefficient other than 1

Solve 2x^{2} + 8x - 10 = 0 .

Divide everything by 2: x^{2} + 4x - 5 = 0
Half of 4 is 2, and 2^{2} = 4 . Add 4 to both sides:
x^{2} + 4x + 4 - 5 = 4
(x + 2)^{2} = 9
x + 2 = \pm 3
⟹ x = 1 or x = −5

Dividing first is essential — the halve-and-square rule assumes the leading coefficient is 1.

Example When there is no real solution

Solve x^{2} + 4x + 8 = 0 .

Half of 4 is 2, and 2^{2} = 4 . Add 4 to both sides:
x^{2} + 4x + 4 + 8 = 4
(x + 2)^{2} = -4
No real number squares to a negative result.
⟹ no real solutions

The method reports the failure clearly, and the negative right-hand side is exactly what a negative discriminant means.

Example A rectangle

A rectangle is 3 units longer than it is wide and has an area of 40 square units. Find its dimensions.

Let the width be x , so the length is x + 3 .
x(x + 3) = 40 , that is x^{2} + 3x - 40 = 0
Half of 3 is 1.5, and 1.5^{2} = 2.25 . Add it to both sides:
(x + 1.5)^{2} = 42.25
x + 1.5 = \pm 6.5 , so x = 5 or x = -8
A width cannot be negative, so the second root is rejected.
⟹ width 5, length 8
Summary
  1. Make the coefficient of x² equal to 1 before doing anything else.
  2. Halve the coefficient of x, square it, and add the result to both sides.
  3. Write the left side as (x + m)² and take the square root, keeping the ± sign.
  4. A negative right-hand side means there are no real solutions.
231. Arithmetic Sequences and Series
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