Lesson Objectives
• Learn how to use the square root property to solve equations such as: x2 = k
• Learn how to solve quadratic equations by completing the square
• Learn how to solve quadratic equations with imaginary solutions

## How to Solve Quadratic Equations by Completing the Square

A quadratic equation has a squared term and no terms of a higher degree.
ax2 + bx + c, a ≠ 0
So far, we have only learned how to solve quadratic equations by factoring.
x2 - 4x + 3 = 0
Factor the left side:
(x - 1)(x - 3) = 0
Use the zero-product property to solve the equation. This means we set each factor with a variable equal to zero and solve the equation.
x - 1 = 0
x = 1
x - 3 = 0
x = 3
x = 1,3
In many cases, we will face quadratic equations that can't be solved by factoring. When this situation occurs, we will generally use the quadratic formula. This formula provides a fast and easy way to obtain a solution for any quadratic equation. Before we jump into the quadratic formula, we need to understand where it comes from and how it works. We will begin by learning about the square root property.

### Square Root Property

If x2 = k, then: $$x=\sqrt{k}\hspace{.5em}or \hspace{.5em}x=-\sqrt{k}$$ This is often shown using the plus or minus symbol to condense the two solutions into one: $$x=\pm \sqrt{k}$$ Let's look at a few examples.
Example 1: Solve each equation using the square root property. $$x^2=16$$ To get x by itself, we will take the square root of each side: $$\sqrt{x^2}=\pm \sqrt{16}$$ $$x=\pm \sqrt{16}$$ $$x=\pm 4$$ The plus or minus in front of the number 4 tells us we have two solutions. x can be 4 or x can be -4, either works as a solution.
As a quick note, notice we can obtain the same solution by factoring. $$x^2=16$$ $$x^2 - 16=0$$ $$(x + 4)(x - 4)=0$$ $$x + 4=0$$ $$x=-4$$ $$x - 4=0$$ $$x=4$$ Either way, we get the same result.
Example 2: Solve each equation using the square root property. $$10x^2 - 1=3239$$ To use the square root property, we need to set up the equation in the form of:
x2 = k
We will add 1 to each side of our equation: $$10x^2=3240$$ We will divide each side by 10: $$x^2=324$$ Now let's solve the equation using the square root property: $$\sqrt{x^2}=\pm \sqrt{324}$$ $$x=\pm \sqrt{324}$$ $$x=\pm 18$$ Example 3: Solve each equation using the square root property. $$2(x-12)^2=98$$ Let's begin by dividing each side by 2: $$(x - 12)^2=49$$ Now, we will use our square root property: $$\sqrt{(x - 12)^2}=\pm \sqrt{49}$$ $$x - 12=\pm 7$$ At this point, we have to solve two equations: $$x - 12=7$$ $$x=19$$ $$x - 12=-7$$ $$x=5$$ $$x=19,5$$

### Solving Quadratic Equations by Completing the Square

What can we do when we encounter a problem such as: $$x^2-10x-80=0$$ We will notice that we can't factor the left side of the equation. There aren't two integers whose sum is -10 and whose product is -80. For this scenario, we can solve the equation by completing the square. Recall that a perfect square is a number whose square root is a rational number. When we have a perfect square trinomial, it is a trinomial that can be factored into a binomial squared.
Perfect Square Trinomial: $$x^2 + 6x + 9=(x + 3)^2$$ $$x^2 - 14x + 49=(x - 7)^2$$ To solve a quadratic equation by completing the square, we will transform the left side of our equation into a perfect square trinomial. We can then factor this into a binomial squared. In this format, we can use the square root property to solve our equation. The process of changing a non-perfect square trinomial into one is known as "completing the square". Let's look at the procedure.
Completing the Square
• Make sure the coefficient of the squared term is 1. If it is not, we will divide each side of the equation by the coefficient of the squared variable.
• Move all variable terms to one side and the constant to the other
• Multiply the coefficient of the first-degree term (raised to the power of 1) by 1/2, then square the result.
• Add the result from the last step to each side of the equation
• Factor the left side of the equation
• Solve the equation using the square root property
Let's look at a few examples.
Example 4: Solve each equation by completing the square. $$x^2 - 14x + 33=0$$ Step 1) Make sure the coefficient of the squared term is 1.
In this case, we have a coefficient of 1 on the squared term.
Step 2) Move all variable terms to one side and the constant to the other. $$x^2 - 14x=-33$$ Step 3) Multiply the coefficient of the first-degree term (raised to the power of 1) by 1/2, then square the result. $$\left(\frac{1}{2}\cdot -14\right)^2=(-7)^2=49$$ Step 4) Add the result from the last step to each side of the equation.
We will add 49 to each side of the equation: $$x^2 - 14x + 49=-33 + 49$$ $$x^2 - 14x + 49=16$$ Step 5) Factor the left side of the equation. $$(x - 7)^2=16$$ Step 6) Solve the equation using the square root property. $$(x - 7)^2=16$$ $$\sqrt{(x - 7)^2}=\pm \sqrt{16}$$ $$x - 7=\pm 4$$ This gives us two equations to solve: $$x - 7=4$$ $$x=11$$ $$x - 7=-4$$ $$x=3$$ $$x=11,3$$ Example 5: Solve each equation by completing the square. $$5x^2+16x+12=0$$ Step 1) Make sure the coefficient of the squared term is 1.
Let's divide each side of the equation by 5, the coefficient of x2. $$x^2 + \frac{16}{5}x + \frac{12}{5}=0$$ Step 2) Move all variable terms to one side and the constant to the other. $$x^2 + \frac{16}{5}x=-\frac{12}{5}$$ Step 3) Multiply the coefficient of the first-degree term (raised to the power of 1) by 1/2, then square the result. The coefficient of our first-degree term is 16/5. $$\left(\frac{16}{5}\cdot \frac{1}{2}\right)^2=\left(\frac{8}{5}\right)^2=\frac{64}{25}$$ Step 4) Add the result from the last step to each side of the equation. $$x^2 + \frac{16}{5}x + \frac{64}{25}=-\frac{12}{5}+ \frac{64}{25}$$ $$x^2 + \frac{16}{5}x + \frac{64}{25}=\frac{4}{25}$$ Step 5) Factor the left side of the equation. $$x^2 + \frac{16}{5}x + \frac{64}{25}=\frac{4}{25}$$ $$\left(x + \frac{8}{5}\right)^2=\frac{4}{25}$$ Step 6) Solve the equation using the square root property. $$\sqrt{\left(x + \frac{8}{5}\right)^2}=\pm \sqrt{\frac{4}{25}}$$ $$x + \frac{8}{5}=\pm \frac{2}{5}$$ This gives us two equations to solve: $$x + \frac{8}{5}=\frac{2}{5}$$ $$x=-\frac{6}{5}$$ $$x + \frac{8}{5}=-\frac{2}{5}$$ $$x=-2$$ $$x=-\frac{6}{5}, -2$$

### Solving Quadratic Equations with Imaginary Solutions

In the last lesson, we learned about complex numbers and the imaginary unit i. In that lesson, we learned the square root of -1 is represented with the imaginary unit i. At this point, we should know the following rule: $$\sqrt{-b}=i\sqrt{b}$$ Let's look at an example that involves a complex solution.
Example 6: Solve each equation by completing the square. $$3x^2 + 6x + 9=0$$ Step 1) Make sure the coefficient of the squared term is 1.
Let's divide each side of the equation by 3, the coefficient of x2. $$x^2 + 2x + 3=0$$ Step 2) Move all variable terms to one side and the constant to the other. $$x^2 + 2x=-3$$ Step 3) Multiply the coefficient of the first-degree term (raised to the power of 1) by 1/2, then square the result. The coefficient of our first-degree term is 2. $$\left(2 \cdot \frac{1}{2}\right)^2=1^2=1$$ Step 4) Add the result from the last step to each side of the equation. $$x^2 + 2x + 1=-3 + 1$$ $$x^2 + 2x + 1=-2$$ Step 5) Factor the left side of the equation. $$(x + 1)^2=-2$$ Step 6) Solve the equation using the square root property. $$\sqrt{(x + 1)^2}=\pm \sqrt{-2}$$ $$x + 1=\pm i\sqrt{2}$$ This gives us two equations to solve: $$x + 1=i\sqrt{2}$$ $$x=-1 + i\sqrt{2}$$ $$x + 1=-i\sqrt{2}$$ $$x=-1 - i\sqrt{2}$$ $$x=-1 \pm i\sqrt{2}$$

#### Skills Check:

Example #1

Solve each equation. $$2x^{2}- 1=119$$

A
$$x=\pm \sqrt{59}$$
B
$$x=\pm 2\sqrt{15}$$
C
$$x=\pm 59$$
D
$$x=\pm \frac{1}{3}$$
E
$$x=\pm \sqrt{13}$$

Example #2

Solve each equation. $$x^{2}+ 4x + 25=0$$

A
$$x=-8 \pm 6i$$
B
$$x=-2 \pm \sqrt{29}$$
C
$$x=-2 \pm i\sqrt{21}$$
D
$$x=-4 \pm i\sqrt{21}$$
E
$$x=-5 \pm i\sqrt{29}$$

Example #3

Solve each equation. $$4x^{2}+ 8x - 21=0$$

A
$$x=-\frac{11}{3}, \frac{5}{3}$$
B
$$x=-13, 19$$
C
$$x=-8 \pm \sqrt{106}$$
D
$$x=-\frac{7}{2}, \frac{3}{2}$$
E
$$x=-4 \pm \sqrt{103}$$