Taking the square root

The square root of a number $$a$$, denoted by $$\sqrt{a}$$, is a number $$b$$ which, when multiplied by itself, gives the number $$a$$. The operation of taking the square root can be essential when solving quadratic equations because this is the way to isolate the squared variable.

Generally, when the equation in the form $$x^n=\text{constant}$$, we can take the $$n$$-th root to solve it!

What does it mean to take the square root?

Taking the square root is a method used for solving quadratic equations in the form $$x^2=\text{constant}$$. It means to calculate the square root of both sides of the equation in order to get the solutions of the equation. This method is used when the squared variable is on the one side and the constant is on the other side. By taking the square root, the square cancels itself with an index of the square root.

Be careful when taking the square root because $$\sqrt{x^2}=|x|$$, so you have two solutions, positive and negative.

‘’Wait…what is a quadratic equation?!’’

Good question! A quadratic equation is an equation in which the variable is raised to the second power.

A quadratic equation written in standard form is:

Why is taking the square root so useful?

Taking the square root is a method for solving a quadratic equation. When solving a quadratic equation by any method, this is the method that is almost always used at the end of the solving. For example, when solving a quadrating equation by completing the square, we solve it by transforming it into the form $$x^2=\text{constant}$$ and then taking the square root.

How to take the square root

Now that we know what the square root is and why it’s useful, it’s time to see it in action! Let’s walk through a problem together.

Solve a quadratic equation by taking the square root:

Take the square root of both sides of the equation and remember to use both positive and negative roots. We can do this because of the rule that states that if two expressions are equal, then their square roots are also equal.

$$x=\pm 3$$

Separate the equation into $$2$$ possible cases (the one with minus root and the one with plus root):

$$x=-3$$

$$x=3$$

Hence, the equation has $$2$$ solutions:

Solve a quadratic equation by taking the square:

Move the constant to the right side of the equation and change its sign:

$$t^2=7+1$$

Add the numbers on the right-hand side of the equation:

$$t^2=8$$

Take the square root of both sides of the equation and remember to use both positive and negative roots. We can do this because of the rule that states that if two expressions are equal, then their square roots are also equal.

$$t=\pm \sqrt{8}$$

Write the number in exponential form with the base of $$2$$:

$$t=\pm \sqrt{2^3}$$

Rewrite the exponent as a sum where one of the addends is a multiple of the index:

$$t=\pm \sqrt{2^{2+1}}$$

Use $$a^{m+n}=a^m\times a^n$$ to expand the expression:

$$t=\pm \sqrt{2^2\times 2^1}$$

Any expression raised to the power of $$1$$ equals itself:

$$t=\pm \sqrt{2^2\times 2}$$

The root of a product is equal to the product of the roots of each factor so rewrite the equation:

$$t=\pm \sqrt{2^2}\times \sqrt{2}$$

Reduce the index of the radical and the exponent with $$2$$:

$$t=\pm 2\sqrt{2}$$

Separate the equation into $$2$$ possible cases (the one with minus root and the one with plus root):

$$t=-2\sqrt{2}$$

$$t=2\sqrt{2}$$

Hence, the equation has $$2$$ solutions:

That wasn’t so bad, right? Now that we’ve walked through a detailed example, let’s review the overall process so you can learn how to use it with any problem:

Do it yourself!

You might not think practicing math is the most glamorous way to spend your time, but the repetition really helps cement the method in your mind. So, when you’re ready, we’ve got some practice problems for you!

Solve a quadratic equation by taking the square root:

1. $$x^2-7=0$$
2. $$-2t^2+15=0$$
3. $$5x^2=25$$
4. $$2a^2-10=20$$

Solutions:

1. $$x_1=-\sqrt{7},x_2=\sqrt{7}$$
2. $$t_1=-\frac{\sqrt{30}}{2},t_2=\frac{\sqrt{30}}{2}$$
3. $$x_1=-\sqrt{5},x_2=\sqrt{5}$$
4. $$a_1=-\sqrt{15},a_2=\sqrt{15}$$

If you’re still struggling through the solving process, that’s totally okay! Stumbling a few times is actually good for learning. If you get too stuck or lost, scan the problem using your Photomath app, and we’ll walk you through to the other side!

Here’s a sneak peek of what you’ll see: