Lesson 13: Completing the Square (Part 2)

Completing the Square (Part 2)

Student 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.

$\displaystyle \begin{aligned} x^2 + 5x - \frac{75}{4}+ \frac{75}{4} &= 0+\frac{75}{4}\\ x^2 + 5x &= \frac{75}{4} \end{aligned}$

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:

$\displaystyle \begin{aligned}x^2 + 5x + \frac{25}{4} &= \frac{75}{4} + \frac{25}{4}\\x^2 + 5x + \frac{25}{4} &= \frac{100}{4} \end{aligned}$

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.

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

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

Visual / Anchor Chart

Standards

Addressing

A-REI.A

HSA-REI.A

A-REI.4.b

HSA-REI.B.4.b

A-REI.4.b

HSA-REI.B.4.b