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
$$x=\pm \frac{3i\sqrt{7}}{2}$$