# Lesson 13

Completing the Square (Part 2)

- Let’s solve some harder quadratic equations.

### 13.1: Math Talk: Equations with Fractions

Solve each equation mentally.

\(x+x=\frac14\)

\((\frac32) ^2 = x\)

\(\frac35 + x = \frac95\)

\(\frac{1}{12}+x=\frac14\)

### 13.2: Solving Some Harder Equations

Solve these equations by completing the square.

- \((x-3)(x+1)=5\)
- \(x^2 + \frac12 x = \frac{3}{16}\)
- \(x^2+3x+\frac84=0\)
- \((7-x)(3-x)+3=0\)
- \(x^2+1.6x+0.63=0\)

- Show that the equation \(x^2+10x+9=0\) is equivalent to \((x+3)^2+4x=0\).
- Write an equation that is equivalent to \(x^2+9x+16=0\) and that includes \((x+4)^2\).
- Does this method help you find solutions to the equations? Explain your reasoning.

### 13.3: Spot Those Errors!

Here are four equations, followed by worked solutions of the equations. Each solution has at least one error.

- Solve one or more of these equations by completing the square.
- Then, look at the worked solution of the same equation as the one you solved. Find and describe the error or errors in the worked solution.

- \(x^2 + 14x= \text-24\)
- \(x^2 - 10x + 16= 0\)
- \(x^2 + 2.4x = \text-0.8\)
- \(x^2 - \frac65 x + \frac15 = 0\)

Worked solutions (with errors):

1.

\(\displaystyle \begin {align} x^2 + 14x &= \text-24\\ x^2 + 14x + 28 &= 4\\ (x+7)^2 &= 4\\ \\x+7 = 2 \quad &\text {or} \quad x+7 = \text-2\\ x = \text-5 \quad &\text {or} \quad x = \text-9 \end{align}\)

2.

\(\displaystyle \begin {align} x^2 - 10x + 16 &= 0\\x^2 - 10x + 25 &= 9\\(x - 5)^2 &= 9\\ \\x-5=9 \quad &\text {or} \quad x-5 = \text-9\\ x=14 \quad &\text {or} \quad x=\text-4 \end{align}\)

3.

\(\displaystyle \begin {align}x^2 + 2.4x &= \text-0.8\\x^2 + 2.4x + 1.44 &= 0.64\\(x + 1.2)^2&=0.64\\x+1.2 &= 0.8\\ x &=\text -0.4 \end{align}\)

4.

\(\displaystyle \begin {align} x^2 - \frac65 x + \frac15 &= 0\\x^2 - \frac65 x + \frac{9}{25} &= \frac{9}{25}\\ \left(x-\frac35\right)^2 &= \frac{9}{25}\\ \\x-\frac35= \frac35 \quad &\text {or} \quad x-\frac35=\text- \frac35\\ x=\frac65 \quad &\text {or} \quad x=0 \end{align}\)

### Summary

Completing the square can be a useful method for solving quadratic equations in cases in which it is not easy to rewrite an expression in factored form. For example, let’s solve this equation:

\(\displaystyle x^2 + 5x - \frac{75}{4}=0\)

First, we’ll add \(\frac{75}{4}\) to each side to make things easier on ourselves.

\(\displaystyle \begin {align} x^2 + 5x - \frac{75}{4}+ \frac{75}{4} &= 0+\frac{75}{4}\\ x^2 + 5x &= \frac{75}{4} \end {align}\)

To complete the square, take \(\frac12\) of the coefficient of the linear term 5, which is \(\frac52\), and square it, which is \(\frac{25}{4}\). Add this to each side:

\(\displaystyle \begin {align}x^2 + 5x + \frac{25}{4} &= \frac{75}{4} + \frac{25}{4}\\x^2 + 5x + \frac{25}{4} &= \frac{100}{4} \end{align}\)

Notice that \(\frac{100}{4}\) is equal to 25 and rewrite it:

\(\displaystyle x^2 + 5x + \frac{25}{4} =25\)

Since the left side is now a perfect square, let’s rewrite it:

\(\displaystyle \left(x+\frac52 \right)^2 = 25\)

For this equation to be true, one of these equations must true:

\(\displaystyle x + \frac52 = 5 \quad \text{or} \quad x + \frac52 = \text-5\)

To finish up, we can subtract \(\frac52\) from each side of the equal sign in each equation.

\(\displaystyle \begin {align} x = 5 - \frac52 \quad &\text{or} \quad x = \text-5 - \frac52\\x = \frac{5}{2} \quad &\text{or} \quad x = \text-\frac{15}{2}\\x = 2\frac12 \quad &\text{or} \quad x = \text-7\frac12 \end{align}\)

It takes some practice to become proficient at completing the square, but it makes it possible to solve many more equations than you could by methods you learned previously.

### Glossary Entries

**completing the square**Completing the square in a quadratic expression means transforming it into the form \(a(x+p)^2-q\), where \(a\), \(p\), and \(q\) are constants.

Completing the square in a quadratic equation means transforming into the form \(a(x+p)^2=q\).

**perfect square**A perfect square is an expression that is something times itself. Usually we are interested in situations where the something is a rational number or an expression with rational coefficients.

**rational number**A rational number is a fraction or the opposite of a fraction. Remember that a fraction is a point on the number line that you get by dividing the unit interval into \(b\) equal parts and finding the point that is \(a\) of them from 0. We can always write a fraction in the form \(\frac{a}{b}\) where \(a\) and \(b\) are whole numbers, with \(b\) not equal to 0, but there are other ways to write them. For example, 0.7 is a fraction because it is the point on the number line you get by dividing the unit interval into 10 equal parts and finding the point that is 7 of those parts away from 0. We can also write this number as \(\frac{7}{10}\).

The numbers \(3\), \(\text-\frac34\), and \(6.7\) are all rational numbers. The numbers \(\pi\) and \(\text-\sqrt{2}\) are not rational numbers, because they cannot be written as fractions or their opposites.