What if for homework you were given a perfect square trinomial to factor, which you wrote in your notebook. However, you accidentally dropped your notebook in a mud puddle, and the last term in the trinomial became obscured. All you can currently see are the first two terms, \begin{align*}x^2 + 16x\end{align*}. How would you go about finding the third term? Once you found the third term, if you factor the trinomial and set it equal to 0, what would be the solution(s) to the equation?

### Completing the Square

**Completing the square** is a method used to create a perfect square trinomial.

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

Most completing the square problems involve finding the third term when you are given the first two terms. In other words, you want to find the missing value in \begin{align*}a^2 + 2(ab) + ?\end{align*}.

We want to find \begin{align*}b\end{align*}. To do this, use the definition of the middle term of the perfect square trinomial and divide that by 2 times the square root of the first term. Once you have \begin{align*}b\end{align*}, you can find \begin{align*}b^2\end{align*}. If equation is in the form \begin{align*}y=x^2+\left(\frac{1}{2} b \right )x+b^2\end{align*} with the coefficient of the first term being 1, then you can simply divide the middle term by 2 and square it to find \begin{align*}b^2\end{align*}.

#### Let's find the missing value to create a perfect square trinomial for the following expressions:

- \begin{align*}x^2+8x+?\end{align*}

The value of \begin{align*}a\end{align*} is \begin{align*}x\end{align*}. To find \begin{align*}b\end{align*}, use the definition of the middle term of the perfect square trinomial.

\begin{align*}&& 2(ab) &= 8x\\ a \ \text{is} \ x, && 2(xb) &= 8x\\ \text{Solve for} \ b: && \frac{2xb}{2x} &= \frac{8x}{2x} \rightarrow b=4\end{align*}

To complete the square you need the value of \begin{align*}b^2\end{align*}.

\begin{align*}b^2=4^2=16\end{align*}

The missing value is 16. The perfect square trinomial we are looking for is \begin{align*}x^2+8x+16\end{align*}.

This is in the form \begin{align*}y=x^2+\left(\frac{1}{2} b \right )x+b^2\end{align*} so you can test the formula. It works because \begin{align*}\frac{1}{2}(8)=4\end{align*} and \begin{align*}4^2=16\end{align*}.

- \begin{align*}x^2+22x+\end{align*}?

We can use the definition of the middle term from \begin{align*}y=x^2+\left(\frac{1}{2} b \right )x+b^2\end{align*} to complete the square.

\begin{align*}\frac{1}{2} (b)=\frac{1}{2} (22)=11\end{align*}

Therefore, \begin{align*}11^2=121\end{align*} and the perfect square trinomial is \begin{align*}x^2+22x+121\end{align*}. Rewriting in its factored form, the equation becomes \begin{align*}(x+11)^2\end{align*}.

#### Solving Equations by Completing the Square

Once you have completed the square, you can solve using square roots.

#### Let's solve the following equations:

- Solve \begin{align*}x^2+22x+121=0\end{align*}.

By completing the square and factoring, the equation becomes:

\begin{align*}&& (x+11)^2 &= 0\\ \text{Solve by taking the square root:} && x+11 &= \pm0\\ \text{Separate into two equations:} && x+11 &=0 \ or \ x+11=0\\ \text{Solve for} \ x: && x &= -11\end{align*}

- Solve \begin{align*}k^2-6k+8=0\end{align*}.

Using the definition to complete the square, \begin{align*}\frac{1}{2}(b)=\frac{1}{2}(-6)=-3\end{align*}. Therefore, the last value of the perfect square trinomial is \begin{align*}(-3)^2=9\end{align*}. The equation given is:

\begin{align*}k^2-6k+8=0, \ and \ 8 \neq 9\end{align*}

In order to factor, complete the square by subtracting 8 and then adding 9 to each side:

\begin{align*}k^2-6k=-8\end{align*}

Remember to use the Addition Property of Equality.

\begin{align*}&& k^2-6k+9 &= -8+9\\ \text{Factor the left side.} && (k-3)^2 &= 1\\ \text{Solve using square roots.} && \sqrt{(k-3)^2} &= \pm \sqrt{1}\\ && k-3 &=1 \ or \ k-3=-1\\ && k &= 4 \ or \ k=2\end{align*}

### Examples

#### Example 1

Earlier, you were told that you have the first two terms of a perfect square trinomial \begin{align*}x^2 + 16x\end{align*} but do not know the third term. How can you find the third term for \begin{align*}x^2 + 16x\end{align*}? Once you found the third term, if you factor the trinomial and set it equal to 0, what are the solutions to the equation?

To find the third term, we can complete the square.

We can use the definition of the middle term from \begin{align*}y=x^2+\left(\frac{1}{2} b \right )x+b^2\end{align*} to complete the square.

\begin{align*}\frac{1}{2} (b)=\frac{1}{2} (16)=8\end{align*}

Therefore, \begin{align*}8^2=64\end{align*} and the perfect square trinomial is \begin{align*}x^2+16x+64\end{align*}. Rewriting in its factored form, the equation becomes \begin{align*}(x+8)^2\end{align*}.

Now, set the factored form of the equation equal to 0 and solve:

\begin{align*}(x+8)^2 &= 0\\ x+8 &= 0\\ x &= -8\end{align*}

The only solution to the equation is -8.

#### Example 2

Solve \begin{align*}x^2+10x+9=0\end{align*}.

First, find \begin{align*}b\end{align*}:

\begin{align*}\frac{1}{2}(b)=\frac{1}{2}(10)=5\end{align*}

Therefore, the last value of the perfect square trinomial is \begin{align*}5^2=25\end{align*}. The equation given is:

\begin{align*}x^2+10x+9=0, \ and \ 9 \neq 25\end{align*}

In order to factor, we must turn the left side into a perfect square trinomial.

Subtract 9:

\begin{align*}x^2+10x=-9\end{align*}

Complete the square: Remember to use the Addition Property of Equality.

\begin{align*}&& x^2+10x+25 &= -9+25\\ \text{Factor the left side.} && (x+5)^2 &= 16\\ \text{Solve using square roots.} && \sqrt{(x+5)^2} &= \pm \sqrt{16}\\ && x+5 &=4 \ or \ x+5=-4\\ && x &= -1 \ or \ x=-9\end{align*}

### Review

- What does it mean to “complete the square”?
- Describe the process used to solve a quadratic equation by completing the square.

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=0\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*}

**Mixed Review**

- A ball dropped from a height of four feet bounces 70% of its previous height. Write the first five terms of this sequence. How high will the ball reach on its \begin{align*}8^{th}\end{align*} bounce?
- Rewrite in standard form: \begin{align*}y=\frac{2}{7} x-11\end{align*}.
- Graph \begin{align*}y=5 \left( \frac{1}{2} \right)^x\end{align*}. Is this exponential growth or decay? What is the growth factor?
- Solve for \begin{align*}r: |3r-4| \le 2\end{align*}.
- Solve for \begin{align*}m:-2m+6=-8(5m+4)\end{align*}.
- Factor \begin{align*}4a^2+36a-40\end{align*}.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 10.5.