Lesson Objectives
• Demonstrate an understanding of the square root property
• Learn how to solve a quadratic equation by completing the square

## How to Solve Quadratic Equations by Completing the Square

In our last lesson, we learned how to use the square root property to solve equations such as: $$x^2=k$$ $$(ax + b)^2=k$$ Recall the square root property tells us: $$x^2=k$$ $$x=\pm \sqrt{k}$$ In most cases, we cannot use the square root property right away. Let's suppose we saw an equation such as: $$x^2 + 20x + 75=0$$ In its current state, we cannot use the square root property. What we need to do is perform a procedure known as completing the square. When we complete the square, we are transforming one side of the equation into a perfect square trinomial. This perfect square trinomial can then be factored into a binomial squared. Once this is done, we can use the square root property to solve our equation.

### Perfect Square Trinomial

$$x^2 + 2xy + y^2=(x + y)^2$$ $$x^2 - 2xy + y^2=(x - y)^2$$

### Completing the Square

• Write the quadratic equation where the terms ax2 and bx are on one side, the constant will be on the other
• Make sure the coefficient (a) for the squared term (ax2) is 1
• If the coefficient (a) is not 1, we will just divide both sides of the equation by (a) the coefficient of the squared term
• Complete the square by adding one-half of the coefficient (b) of the first-degree term (bx) squared to both sides of the equation
• Solve the equation using the square root property
Let's revisit our earlier example: $$x^2 + 20x + 75=0$$ Step 1) We want our ax2 and bx terms on one side and the constant on the other.
We will subtract 75 away from each side of the equation: $$x^2 + 20x=-75$$ Step 2) Make sure the coefficient for the squared term is 1.
In this case, we have x2, which is the same as 1x2 $$x^2 + 20x=-75$$ Step 3) Complete the square » add one-half of the coefficient of the first-degree term squared to both sides.
Our first-degree term is 20x, therefore, the coefficient is 20. We want to multiply this number by 1/2 or divide by 2: $$20 \cdot \frac{1}{2}=10$$ Now square the result (10): $$10^2=100$$ Note we can do this in one step: $$\left(20 \cdot \frac{1}{2}\right)^2=10^2=100$$ Now we add this value of 100 to both sides of the equation: $$x^2 + 20x + 100=-75 + 100$$ Simplify: $$x^2 + 20x + 100=25$$ Step 4) Solve the equation.
The left side of the equation is now a perfect square trinomial. This means we can factor the trinomial into a binomial squared: $$x^2 + 20x + 100=25$$ $$(x + 10)^2=25$$ Now we can use our square root property to solve the equation: $$\sqrt{(x + 10)^2}=\pm \sqrt{25}$$ $$x + 10=\pm 5$$ This leads to two equations to solve: $$x + 10=5$$ $$x=-5$$ $$x + 10=-5$$ $$x=-15$$ Our two solutions: $$x=-5 \hspace{.5em}\text{or} \hspace{.5em}x=-15$$ Let's look at a few more examples.
Example 1: Solve each equation. $$4x^2 - 16x - 20=0$$ Step 1) We want our ax2 and bx terms on one side and the constant on the other.
We will add 20 to each side of the equation: $$4x^2 - 16x=20$$ Step 2) Make sure the coefficient for the squared term is 1.
In this case, we have 4x2, which means we have a coefficient of 4. We will divide each part of the equation by 4: $$\frac{4}{4}x^2 - \frac{16}{4}x=\frac{20}{4}$$ $$x^2 - 4x=5$$ Step 3) Complete the square » add one-half of the coefficient of the first-degree term squared to both sides.
Our first-degree term is 4x, therefore, the coefficient is 4. We want to multiply this number by 1/2 or divide by 2: $$4 \cdot \frac{1}{2}=2$$ Now square the result (2): $$2^2=4$$ Now we add this value of 4 to both sides of the equation: $$x^2 - 4x + 4=5 + 4$$ Simplify: $$x^2 - 4x + 4=9$$ Step 4) Solve the equation.
The left side of the equation is now a perfect square trinomial. This means we can factor the trinomial into a binomial squared: $$x^2 - 4x + 4=9$$ $$(x - 2)^2=9$$ Now we can use our square root property to solve the equation: $$\sqrt{(x - 2)^2}=\pm \sqrt{9}$$ $$x - 2=\pm 3$$ This leads to two equations to solve: $$x - 2=3$$ $$x=5$$ $$x - 2=-3$$ $$x=-1$$ Our two solutions: $$x=5 \hspace{.5em}\text{or} \hspace{.5em}x=-1$$ Example 2: Solve each equation. $$8x^2 + 16x - 90=0$$ Step 1) We want our ax2 and bx terms on one side and the constant on the other.
We will add 90 to each side of the equation: $$8x^2 + 16x=90$$ Step 2) Make sure the coefficient for the squared term is 1.
In this case, we have 8x2, which means we have a coefficient of 8. We will divide each part of the equation by 8: $$\frac{8}{8}x^2 + \frac{16}{8}x=\frac{90}{8}$$ $$x^2 + 2x=\frac{45}{4}$$ Step 3) Complete the square » add one-half of the coefficient of the first-degree term squared to both sides.
Our first-degree term is 2x, therefore, the coefficient is 2. We want to multiply this number by 1/2 or divide by 2: $$2 \cdot \frac{1}{2}=1$$ Now square the result (2): $$1^2=1$$ Now we add this value of 1 to both sides of the equation: $$x^2 + 2x + 1=\frac{45}{4}+ 1$$ Simplify: $$x^2 + 2x + 1=\frac{45}{4}+ \frac{4}{4}$$ $$x^2 + 2x + 1=\frac{49}{4}$$ Step 4) Solve the equation.
The left side of the equation is now a perfect square trinomial. This means we can factor the trinomial into a binomial squared: $$x^2 + 2x + 1=\frac{49}{4}$$ $$(x + 1)^2=\frac{49}{4}$$ Now we can use our square root property to solve the equation: $$\sqrt{(x + 1)^2}=\pm \sqrt{\frac{49}{4}}$$ $$x + 1=\pm \frac{7}{2}$$ This leads to two equations to solve: $$x + 1=\frac{7}{2}$$ $$x=\frac{7}{2}- 1$$ $$x=\frac{7}{2}- \frac{2}{2}$$ $$x=\frac{5}{2}$$ $$x + 1=-\frac{7}{2}$$ $$x=-\frac{7}{2}- 1$$ $$x=\frac{-7}{2}- \frac{2}{2}$$ $$x=\frac{-9}{2}$$ $$x=-\frac{9}{2}$$ Our two solutions: $$x=\frac{5}{2}\hspace{.5em}\text{or} \hspace{.5em}x=-\frac{9}{2}$$ Example 3: Solve each equation. $$4x^2 + 13x - 20=-8$$ Step 1) We want our ax2 and bx terms on one side and the constant on the other.
We will add 20 to each side of the equation: $$4x^2 + 13x=12$$ Step 2) Make sure the coefficient for the squared term is 1.
In this case, we have 4x2, which means we have a coefficient of 4. We will divide each part of the equation by 4: $$\frac{4}{4}x^2 + \frac{13}{4}x=\frac{12}{4}$$ $$x^2 + \frac{13}{4}x=3$$ Step 3) Complete the square » add one-half of the coefficient of the first-degree term squared to both sides.
Our first-degree term is (13/4)x, therefore, the coefficient is 13/4. We want to multiply this number by 1/2 or divide by 2: $$\frac{13}{4}\cdot \frac{1}{2}=\frac{13}{8}$$ Now square the result (2): $$\left(\frac{13}{8}\right)^2=\frac{169}{64}$$ Now we add this value of 169/64 to both sides of the equation: $$x^2 + \frac{13}{4}x + \frac{169}{64}=3 + \frac{169}{64}$$ Simplify: $$x^2 + \frac{13}{4}x + \frac{169}{64}=\frac{192}{64}+ \frac{169}{64}$$ $$x^2 + \frac{13}{4}x + \frac{169}{64}=\frac{361}{64}$$ Step 4) Solve the equation.
The left side of the equation is now a perfect square trinomial. This means we can factor the trinomial into a binomial squared: $$x^2 + \frac{13}{4}x + \frac{169}{64}=\frac{361}{64}$$ $$\left(x + \frac{13}{8}\right)^2=\frac{361}{64}$$ Now we can use our square root property to solve the equation: $$\sqrt{\left(x + \frac{13}{8}\right)^2}=\pm \sqrt{\frac{361}{64}}$$ $$x + \frac{13}{8}=\pm \frac{19}{8}$$ This leads to two equations to solve: $$x + \frac{13}{8}=\frac{19}{8}$$ $$x=\frac{19}{8}- \frac{13}{8}$$ $$x=\frac{6}{8}$$ $$x=\frac{3}{4}$$ $$x + \frac{13}{8}=-\frac{19}{8}$$ $$x=-\frac{19}{8}- \frac{13}{8}$$ $$x=-\frac{32}{8}$$ $$x=-4$$ Our two solutions: $$x=\frac{3}{4}\hspace{.5em}\text{or} \hspace{.5em}x=-4$$

#### Skills Check:

Example #1

Solve each equation. $$x^2=-48 + 6x$$

A
$$x=-\frac{3}{7}, 7$$
B
$$x=-2, 9$$
C
$$x=3 \pm \sqrt{57}$$
D
$$x=6 \pm 2i \sqrt{3}$$
E
$$x=3 \pm i\sqrt{39}$$

Example #2

Solve each equation. $$9x^2=18x - 8$$

A
$$x=-11, 16$$
B
$$x=\frac{1}{3}, \frac{8}{7}$$
C
$$x=\frac{2}{3}, \frac{4}{3}$$
D
$$x=\pm 2i\sqrt{5}$$
E
$$x=3 \pm 6i$$

Example #3

Solve each equation. $$4x^2 + 8x=77$$

A
$$x=-2, 6$$
B
$$x=-1, 19$$
C
$$x=-\frac{11}{2}, \frac{7}{2}$$
D
$$x=-4, \frac{2}{5}$$
E
$$x=\pm \frac{3i\sqrt{7}}{2}$$         