Completing the Square

Completing the square rewrites a quadratic using a squared binomial. It is both a solving method and a way to reveal the vertex and extreme value of a quadratic function.

The key pattern is $x^{2} + bx$ . Adding $(\frac{b}{2})^{2}$ creates the square $(x + \frac{b}{2})^{2}$ .

On this page (8 sections)

The perfect-square pattern

Expanding $(x+r)^2$ gives $x^2+2rx+r^2$ . To match a middle term $bx$ , choose $r=b/2$ . Adding and subtracting the same quantity keeps the expression unchanged.

x^2+bx=\left(x+\frac b2\right)^2-\left(\frac b2\right)^2

The identity holds for real $x$ and $b$ . It is not permission to add a new constant without compensation: an expression must keep the same value, and an equation must receive the same addition on both sides.

Rewriting an expression

Rewrite $x^2+6x+5$

$x^2+6x+5=x^2+6x+9-9+5$
Half of 6 is 3, and its square is 9. Add and subtract 9.
$(x+3)^2-4$
Group the perfect square and combine the constants.

Completing a square with a 3-by-3 corner

x

3

x

3

x^2

3x

3x

9

added corner

Adding the corner gives

x^2+6x+9=(x+3)^2

Diagram explained

For positive x, start with an x-by-x square and two strips each of width 3. The unfilled corner is 3 by 3. In an expression, compensate for this added area by subtracting 9; in an equation, add 9 to both sides.

In the area model, $x$ is a positive length. Adding a $3\times3$ corner completes a square of side $x+3$ . The algebraic identity remains valid for negative $x$ as well, even though those values are not side lengths in the model.

Solving by completing the square

Solve $x^2+6x-7=0$

$x^2+6x=7$
Add 7 to both sides.
$x^2+6x+9=16$
Add 9 to both sides, not just the left.
$(x+3)^2=16$
Recognize the perfect square.
$x+3=\pm4$
Use both signs.
$x=-3\pm4$
Subtract 3.

Therefore $x=1$ or $x=-7$ . Substituting into the original quadratic gives zero in both cases.

When the leading coefficient is not 1

When the leading coefficient is not 1, factor it from the quadratic and linear terms before completing the square.

2x^2+8x+3=2(x^2+4x)+3

=2\bigl((x+2)^2-4\bigr)+3

=2(x+2)^2-5

The compensation $-4$ is inside the factor $2$ , so it contributes $-8$ to the entire expression. Subtracting only $4$ outside would be incorrect.

Expanding checks the result: $2(x+2)^2-5=2x^2+8x+8-5=2x^2+8x+3$ .

Finding a quadratic’s vertex and minimum

For $y=2(x+2)^2-5$ , the non-negative square shows that the minimum output is $-5$ , attained at $x=-2$ . The vertex is $(-2,-5)$ . Completing the square can therefore answer a minimum-value question as well as solve an equation.

Applying the same method to a general quadratic produces the quadratic formula. It also gives the vertex form used when describing transformations of a parabola.

Practice questions

Rewrite $x^{2} + 6x + 5$ by completing the square.
Solve $x^{2} + 6x – 7 = 0$ .
What number completes $x^{2} + 10x$ to a perfect square?

Check your answers

Add and subtract 9: $x^2+6x+5=(x+3)^2-4$ .
$(x+3)^2=16$ , so $x+3=\pm4$ and $x=1$ or $x=-7$ .
$25$ , because $(10/2)^2=25$ and $(x+5)^2=x^2+10x+25$ .

Sources and further reading

OpenStax: Vertex form and completing the square. Definitions, conventions and related examples.
OpenStax Elementary Algebra 2e. Language of algebra, equations, inequalities and factoring.
OpenStax Intermediate Algebra 2e. Quadratic equations, factoring and algebraic techniques.
Wolfram Language: Algebraic Transformations. Symbolic simplification, expansion and factoring reference.