Math Core

Square roots solve $(x + 3)^2 = 16$ in two lines. But most quadratics, like $x^2 + 6x - 7 = 0$ , don't arrive as a perfect square. Completing the square is a way to rewrite any quadratic so that it does, which means you can then solve it with square roots, even when it won't factor. The same move also converts standard form into vertex form.

Perfect square trinomials

In the special products lesson you saw that squaring a binomial follows a pattern:

(x + p)^2 = x^2 + 2px + p^2.

Look at the relationship between the middle coefficient $2p$ and the constant $p^2$ : the constant is half the middle coefficient, squared.

trinomial	half of $b$	$\left(\frac{b}{2}\right)^2$	as a square
$x^2 + 6x + 9$	$3$	$9$	$(x + 3)^2$
$x^2 - 10x + 25$	$-5$	$25$	$(x - 5)^2$
$x^2 + 3x + \frac{9}{4}$	$\frac{3}{2}$	$\frac{9}{4}$	$\left(x + \frac{3}{2}\right)^2$

A picture explains the name. Think of $x^2 + 6x$ as an $x$ -by- $x$ square plus a strip of area $6x$ . Cut the strip in half and attach one $3$ -by- $x$ piece to the right side and one to the bottom. The shape is almost a big square with side $x + 3$ , but a $3$ -by- $3$ corner is missing. Adding that corner, $9$ , completes the square.

Completing the square

To make $x^2 + bx$ into a perfect square, add $\left(\dfrac{b}{2}\right)^2$ :

x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2.

This works when the coefficient of $x^2$ is $1$ .

Worked example: Finding the missing constant

What number completes the square? Write the result as a binomial squared.

$x^2 - 14x$
$x^2 + 5x$

Solutions.

Half of $-14$ is $-7$ , and $(-7)^2 = 49$ . So $x^2 - 14x + 49 = (x - 7)^2$ .
Half of $5$ is $\dfrac{5}{2}$ , and $\left(\dfrac{5}{2}\right)^2 = \dfrac{25}{4}$ . So $x^2 + 5x + \dfrac{25}{4} = \left(x + \dfrac{5}{2}\right)^2$ .

Solving by completing the square

Move the constant term to the right side.
If the coefficient of $x^2$ isn't $1$ , divide every term by it.
Add $\left(\dfrac{b}{2}\right)^2$ to both sides.
Write the left side as a binomial squared.
Take square roots (with $\pm$ ) and solve.

Worked example: A first example

Solve $x^2 + 6x - 7 = 0$ by completing the square.

\begin{aligned} x^2 + 6x &= 7 && \text{move the constant} \\ x^2 + 6x + 9 &= 7 + 9 && \text{add } \left(\tfrac{6}{2}\right)^2 = 9 \text{ to both sides} \\ (x + 3)^2 &= 16 \\ x + 3 &= \pm 4 \\ x = 1 \quad &\text{or} \quad x = -7 \end{aligned}

This one also factors as $(x + 7)(x - 1) = 0$ , which confirms the answers.

Common mistake

Whatever you add to the left side, you must add to the right side too. Writing $x^2 + 6x + 9 = 7$ changes the equation and gives wrong answers. Think of it as a balance: $+9$ on one side needs $+9$ on the other.

Worked example: An irrational answer

Solve $x^2 - 4x - 3 = 0$ .

This doesn't factor over the integers, but completing the square still works:

\begin{aligned} x^2 - 4x &= 3 \\ x^2 - 4x + 4 &= 3 + 4 && \left(\tfrac{-4}{2}\right)^2 = 4 \\ (x - 2)^2 &= 7 \\ x - 2 &= \pm\sqrt{7} \\ x &= 2 \pm \sqrt{7} \end{aligned}

The solutions are about $4.65$ and $-0.65$ .

Worked example: When a is not 1

Solve $2x^2 + 8x - 5 = 0$ .

Divide every term by $2$ first so the $x^2$ coefficient is $1$ :

\begin{aligned} x^2 + 4x - \tfrac{5}{2} &= 0 \\ x^2 + 4x &= \tfrac{5}{2} \\ x^2 + 4x + 4 &= \tfrac{5}{2} + 4 \\ (x + 2)^2 &= \tfrac{13}{2} \\ x &= -2 \pm \sqrt{\tfrac{13}{2}} \end{aligned}

To tidy the radical, $\sqrt{\dfrac{13}{2}} = \dfrac{\sqrt{13}}{\sqrt{2}} = \dfrac{\sqrt{26}}{2}$ , so $x = -2 \pm \dfrac{\sqrt{26}}{2}$ , about $0.55$ and $-4.55$ .

Vertex form by completing the square

Completing the square on an expression instead of an equation turns standard form into vertex form. Since there's no other side to balance, you add $\left(\frac{b}{2}\right)^2$ and immediately subtract it, which adds zero overall.

Rewrite $y = x^2 + 8x + 10$ :

\begin{aligned} y &= (x^2 + 8x + 16) - 16 + 10 && \text{add and subtract } \left(\tfrac{8}{2}\right)^2 = 16 \\ &= (x + 4)^2 - 6 \end{aligned}

The vertex is $(-4, -6)$ . Setting $y = 0$ gives $(x + 4)^2 = 6$ , so the x-intercepts are $-4 \pm \sqrt{6}$ , about $-1.55$ and $-6.45$ .

y = x² + 8x + 10 = (x + 4)² - 6 has vertex (-4, -6) and x-intercepts -4 ± √6.Open in grapher →

When $a \ne 1$ , factor $a$ out of the $x$ -terms first. For $y = 2x^2 - 12x + 13$ :

\begin{aligned} y &= 2(x^2 - 6x) + 13 \\ &= 2(x^2 - 6x + 9) - 18 + 13 && \text{the added } 9 \text{ is really } 2 \cdot 9 = 18 \\ &= 2(x - 3)^2 - 5 \end{aligned}

The vertex is $(3, -5)$ , which you can confirm with $x = -\dfrac{b}{2a} = \dfrac{12}{4} = 3$ .

Tip

When $a \ne 1$ , the number you add inside the parentheses gets multiplied by $a$ . Subtract $a \cdot \left(\frac{b}{2}\right)^2$ outside, not just $\left(\frac{b}{2}\right)^2$ . Expanding your final answer is a quick check.

Practice

Practice 1

What number must be added to $x^2 + 18x$ to make a perfect square trinomial?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Find $c$ so that $x^2 - 3x + c$ is a perfect square trinomial.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Solve $x^2 + 10x + 16 = 0$ by completing the square.

Separate answers with commas, e.g. 2, -5

Practice 4

Solve $x^2 - 8x + 3 = 0$ by completing the square. What is the larger solution? Give an exact answer, such as 2 + sqrt(5).

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Solve $x^2 + 2x - 5 = 0$ .

Practice 6

Complete the square to write $y = x^2 - 10x + 21$ in vertex form. What is the vertex?

Enter a point like (2, -3)

Practice 7

Complete the square to write $y = 3x^2 + 6x - 2$ in the form $y = a(x - h)^2 + k$ . What is $k$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

Solve $2x^2 - 12x + 7 = 0$ by completing the square. What is the larger solution? Give an exact answer.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.