## Solving Quadratic Equations by Extracting the Square Root

In the previous post, we have learned about quadratic equations or equations of the form $ax^2 + bx + c = 0$, where a is not equal to 0. In this equation, we want to find the value of x which we call the root or the solution to the equation.

There are three strategies in finding the root of the equation: by extracting the roots, by completing the square, and by the quadratic formula. In this example, we will discuss, how to find the root of the quadratic equation by extracting the root.

Just like in solving equations, if we want to find the value of x, we put all the numbers on one side, and all the x’s on one side. Since quadratic equations contain the term $x^2$, we can find the value of x by extracting the square roots. Below are five examples on how to do this.

Example 1: $2x^2 = 0$

Solution

Dividing both sides by 2, we have $\frac{2x^2}{2} = \frac{0}{2}$.

This gives us $x^2 = 0$.

Extracting the square root of both sides, we have $x = 0$.

Therefore, the root $x = 0$.

Example 2: $x^2 - 36 = 0$

Solution $x^2 - 36 = 0$.

Adding 36 to both side, we have $x^2 - 36 + 36 = 0 + 36$. $x^2 = 36$.

Extracting the square root of both sides, we have $\sqrt{x^2} = \sqrt{36}$. $x = \pm 6$.

In this example, x has two roots: x = 6 and x = -6.

Example 3: $x^2 + 81 = 0$

Solution

Subtracting 81 from both sides, we have $x^2 + 81 - 81 = 0 - 81$ $x^2 = -81$ $\sqrt{x^2} = - 81$ $x = \sqrt{-81}$.

In this case, there is no number that when multiplied by itself is negative. For example, negative times negative is equal to positive, and positive times positive is equal to positive. Therefore, there is no real root. There is, however, what we call a complex root as shown in the video below.

Example 4: $5x^2 = 12$

Solution $5x^2 = 12$

Dividing both sides by 5, we have $x^2 = \frac{12}{5}$.

Extracting the square root of both sides, we have $\sqrt{x^2} = \sqrt{\frac{12}{5}} = \frac{\sqrt{12}}{\sqrt{5}}$ $x = \frac{\sqrt{4}\sqrt{3}}{\sqrt{5}}$ $x = \frac{2 \sqrt{3}}{\sqrt{5}}$.

Example 5: $3x^2 = - 4$ $3x^2 = -4$

Dividing both sides by 3, we have $x^2 = \frac{-4}{3}$.

Extracting the square root of both sides, we have $\sqrt{x^2} = \sqrt{\frac{-4}{3}}$.

Again, the sign of the number inside the radical is negative, so there is no real root. To know how to compute for the complex root, watch the video below.

### 3 Comments

• ### zandrah angela palma

July 14, 2019

hey um in the 3rd example why is it that x is equal to the square root of negative 81? i thought it should be 9. please explain it to me because im very confused.

• ### Civil Service Reviewer

August 3, 2019

Because negative numbers has no square roots.

• ### Anriel Alban

September 7, 2019

Can i ask a question

Uhhm not all the time Answer in Factoring and Answer in Quadratic Formula are not the same?

Please enlighten me

Thanks in advance