# 10.6: Completing the Square

**At Grade**Created by: CK-12

**Practice**Completing the Square

What if you had a quadratic equation like \begin{align*}x^2 + 12x = 13\end{align*}

### Watch This

CK-12 Foundation: 1006S Solving Quadratic Equations by Completing the Square

### Guidance

You saw in the last section that if you have a quadratic equation of the form \begin{align*}(x - 2)^2 = 5\end{align*}

\begin{align*}x - 2 = \sqrt{5} \ \qquad \text{and} \qquad \ x - 2 = - \sqrt{5}\end{align*}

Simplify to get:

\begin{align*}x = 2 + \sqrt{5} \approx 4.24 \ \qquad \text{and} \qquad \ x = 2 - \sqrt{5} \approx - 0.24\end{align*}

So what do you do with an equation that isn’t written in this nice form? In this section, you’ll learn how to rewrite any quadratic equation in this form by **completing the square.**

**Complete the Square of a Quadratic Expression**

Completing the square lets you rewrite a quadratic expression so that it contains a perfect square trinomial that you can factor as the square of a binomial.

Remember that the square of a binomial takes one of the following forms:

\begin{align*}(x + a)^2 & = x^2 + 2ax + a^2\\
(x - a)^2 & = x^2 - 2ax + a^2\end{align*}

So in order to have a perfect square trinomial, we need two terms that are perfect squares and one term that is twice the product of the square roots of the other terms.

#### Example A

*Complete the square for the quadratic expression* \begin{align*}x^2 + 4x\end{align*}

**Solution**

To complete the square we need a constant term that turns the expression into a perfect square trinomial. Since the middle term in a perfect square trinomial is always 2 times the product of the square roots of the other two terms, we re-write our expression as:

\begin{align*}x^2 + 2(2)(x)\end{align*}

We see that the constant we are seeking must be \begin{align*}2^2\end{align*}

\begin{align*}x^2 + 2(2)(x) + 2^2\end{align*}

**Answer:** By adding 4 to both sides, this can be factored as: \begin{align*}(x + 2)^2\end{align*}

Notice, though, that we just changed the value of the whole expression by adding 4 to it. If it had been an equation, we would have needed to add 4 to the other side as well to make up for this.

Also, this was a relatively easy example because \begin{align*}a\end{align*}, the coefficient of the \begin{align*}x^2\end{align*} term, was 1. When that coefficient doesn’t equal 1, we have to factor it out from the whole expression before completing the square.

#### Example B

*Complete the square for the quadratic expression* \begin{align*}4x^2 + 32x\end{align*}.

**Solution**

\begin{align*}& \text{Factor the coefficient of the} \ x^2 \ \text{term}: & & 4(x^2 + 8x)\\ & \text{Now complete the square of the expression in parentheses}.\\ & \text{Re-write the expression}: & & 4 (x^2 + 2(4) (x))\\ & \text{We complete the square by adding the constant} \ 4^2: & & 4(x^2 + 2(4)(x) + 4^2)\\ & \text{Factor the perfect square trinomial inside the parenthesis}: & & 4(x + 4)^2 \qquad \text{Answer}\end{align*}

The expression **“completing the square”** comes from a geometric interpretation of this situation. Let’s revisit the quadratic expression in Example 1: \begin{align*}x^2 + 4x\end{align*}.

We can think of this expression as the sum of three areas. The first term represents the area of a square of side \begin{align*}x\end{align*}. The second expression represents the areas of two rectangles with a length of 2 and a width of \begin{align*}x\end{align*}:

We can combine these shapes as follows:

We obtain a square that is not quite complete. To complete the square, we need to add a smaller square of side length 2.

We end up with a square of side length \begin{align*}(x + 2)\end{align*}; its area is therefore \begin{align*}(x + 2)^2\end{align*}. Let’s demonstrate the method of **completing the square** with an example.

#### Example C

*Solve the following quadratic equation:* \begin{align*}3x^2 - 10x = -1\end{align*}

**Solution**

Divide all terms by the coefficient of the \begin{align*}x^2\end{align*} term:

\begin{align*}x^2 - \frac{10}{3} x = - \frac{1}{3}\end{align*}

Rewrite: \begin{align*}x^2 - 2 \left ( \frac{5}{3} \right ) (x) = - \frac{1}{3}\end{align*}In order to have a perfect square trinomial on the right-hand-side we need to add the constant \begin{align*}\left ( \frac{5}{3} \right )^2\end{align*}. Add this constant to **both** sides of the equation:

\begin{align*}x^2 - 2 \left ( \frac{5}{3} \right ) (x) + \left ( \frac{5}{3} \right )^2 = - \frac{1}{3} + \left ( \frac{5}{3} \right )^2\end{align*}

Factor the perfect square trinomial and simplify:

\begin{align*}\left ( x - \frac{5}{3} \right )^2 & = - \frac{1}{3} + \frac{25}{9}\\ \left (x - \frac{5}{3} \right )^2 & = \frac{22}{9}\end{align*}

Take the square root of both sides:

\begin{align*}x - \frac{5}{3} &= \sqrt{\frac{22}{9}} && \text{and} && x - \frac{5}{3} = - \sqrt{ \frac{22}{9}}\\ x &= \frac{5}{3} + \sqrt{\frac{22}{9}} \approx 3.23 && \text{and} && x = \frac{5}{3} - \sqrt{\frac{22}{9}} \approx 0.1\end{align*}

**Answer:** \begin{align*}x = 3.23\end{align*} and \begin{align*}x = 0.1\end{align*}

**Solving Quadratic Equations in Standard Form**

If an equation is in standard form \begin{align*}(ax^2 + bx + c = 0)\end{align*}, we can still solve it by the method of completing the square. All we have to do is start by moving the constant term to the right-hand-side of the equation.

#### Example D

*Solve the following quadratic equation:* \begin{align*}x^2 + 15x + 12 = 0\end{align*}

**Solution**

Move the constant to the other side of the equation:

\begin{align*}x^2 + 15x = -12\end{align*}

Rewrite: \begin{align*}x^2 + 2 \left ( \frac{15}{2} \right )(x) = -12\end{align*}

Add the constant \begin{align*}\left ( \frac{15}{2} \right )^2\end{align*} to both sides of the equation:

\begin{align*}x^2 + 2\left ( \frac{15}{2} \right )(x) + \left ( \frac{15}{2} \right )^2 = -12 + \left ( \frac{15}{2} \right )^2\end{align*}

Factor the perfect square trinomial and simplify:

\begin{align*}\left (x + \frac{15}{2} \right )^2 & = -12 + \frac{225}{4}\\ \left (x + \frac{15}{2} \right )^2 & = \frac{177}{4}\end{align*}

Take the square root of both sides:

\begin{align*}x + \frac{15}{2} &= \sqrt{\frac{177}{4}} && \text{and} && x + \frac{15}{2} = - \sqrt{\frac{177}{4}}\\ x &= - \frac{15}{2} + \sqrt{\frac{177}{4}} \approx - 0.85 && \text{and} && x = - \frac{15}{2} - \sqrt{\frac{177}{4}} \approx -14.15\end{align*}

**Answer:** \begin{align*}x = -0.85\end{align*} and \begin{align*}x = -14.15\end{align*}

Watch this video for help with the Examples above.

CK-12 Foundation: 1006 Solving Quadratic Equations by Completing the Square

### Vocabulary

- A
has the form \begin{align*}a^2+2(ab)+b^2\end{align*}, which factors into \begin{align*}(a+b)^2\end{align*}.*perfect square trinomial*

### Guided Practice

*Solve the following quadratic equation:* \begin{align*}-x^2 +22x = 5\end{align*}

**Solution**

Divide all terms by the coefficient of the \begin{align*}x^2\end{align*} term:

\begin{align*}x^2 -22x = -6\end{align*}

Rewrite: \begin{align*}x^2 - 2(11) (x) = - 6\end{align*}In order to have a perfect square trinomial on the right-hand-side we need to add the constant \begin{align*}\left (11 \right )^2\end{align*}. Add this constant to **both** sides of the equation:

\begin{align*}x^2 - 2 ( 11 ) (x) + ( 11)^2 = - 6 + (11)^2\end{align*}

Factor the perfect square trinomial and simplify:

\begin{align*}\left ( x - 11 \right )^2 & =- 6 + (11)^2 \\ \left (x - \frac{5}{3} \right )^2 & = 16\end{align*}

Take the square root of both sides:

\begin{align*}x - 11 &= \sqrt{16} && \text{and} && x - 11 = - \sqrt{ 16}\\ x &= 11 + \sqrt{16} =15 && \text{and} && x =11 - \sqrt{4}= 7\end{align*}

**Answer:** \begin{align*}x = 15\end{align*} and \begin{align*}x =7\end{align*}

### Practice

Complete the square for each expression.

- \begin{align*}x^2 + 5x\end{align*}
- \begin{align*}x^2 - 2x\end{align*}
- \begin{align*}x^2 + 3x\end{align*}
- \begin{align*}x^2 - 4x\end{align*}
- \begin{align*}3x^2 + 18x\end{align*}
- \begin{align*}2x^2 - 22x\end{align*}
- \begin{align*}8x^2 - 10x\end{align*}
- \begin{align*}5x^2 + 12x\end{align*}

Solve each quadratic equation by completing the square.

- \begin{align*}x^2 - 4x = 5\end{align*}
- \begin{align*}x^2 - 5x = 10\end{align*}
- \begin{align*}x^2 + 10x + 15 = 0\end{align*}
- \begin{align*}x^2 + 15x + 20 = 0\end{align*}
- \begin{align*}2x^2 - 18x = 3\end{align*}
- \begin{align*}4x^2 + 5x = -1\end{align*}
- \begin{align*}10x^2 - 30x - 8 = 0\end{align*}
- \begin{align*}5x^2 + 15x - 40 = 0\end{align*}

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Perfect Square Trinomial

A perfect square trinomial is a quadratic expression of the form (which can be rewritten as ) or (which can be rewritten as ).Square Root

The square root of a term is a value that must be multiplied by itself to equal the specified term. The square root of 9 is 3, since 3 * 3 = 9.### Image Attributions

Here you'll learn how to complete the square to help you solve quadratic equations. You'll also solve quadratic equations in standard form.