Square roots and radicals

Radicals are the opposite of exponents. Instead of multiplying a number by itself multiple times, a radical is a number that must be multiplied by itself several times to get a certain value. For example, $\sqrt{100}$ asks for the number that can be multiplied by itself twice to get $100$, and the answer is $10$ because $10\ast 10=100$.

There are two rules you should keep in mind when you are working with radicals.

Rule 1: Use multiplication and division to combine or split inside radicals

For example:

$$\begin{array}{rl}\sqrt{2}\ast \sqrt{3}& =\sqrt{6}\\ \sqrt{2}\ast \sqrt{3}& =\sqrt{2\ast 3}\\ 2\ast 3& =2\ast 3\\ 2\ast 3& =6\end{array}$$

The same works for division:

$$\begin{array}{rl}\sqrt{12}/\sqrt{6}& =\sqrt{2}\\ \sqrt{12}/\sqrt{6}& =\sqrt{12/6}\\ 12/6& =12/6\\ 12/6& =2\end{array}$$

Rule 2: Treat radicals like variables from the outside

For example:

$$\begin{array}{rl}3\sqrt{2}+2\sqrt{2}& =5\sqrt{2}\end{array}$$

Imagine if $x$ was used to represent the radical $\sqrt{2}$:

$$\begin{array}{rl}3x+2x& =5x\\ 3\sqrt{2}+2\sqrt{2}& =5\sqrt{2}\end{array}$$

It can be useful to translate radicals into exponents and use exponent rules to solve for the answer. Every radical can be described as a fraction exponent. For example:

$$\sqrt{25}=\sqrt[2]{25^1}=25^{(1/2)}$$

The numerator of the exponent is the exponent inside the radical. We can represent 25 with an implied exponent of 1, as anything raised to the power of 1 is itself. All square roots have an implied 2 outside of the radical because a square root asks for the number that, when multiplied by itself, twice gives you the value inside the radical. If there were a 3 outside the radical, it would be a cube root.

Can you rewrite this radical into exponents?

$$\sqrt[3]{12^2}\ast \sqrt[3]{12}$$

All right!

Here’s a more complex question that requires the use of both rules:

Simplify the following equation:

$5\sqrt{3}\ast \sqrt{27}-2\sqrt{81}-\sqrt{18}$

The trick to solving this question is understanding that we can combine $\sqrt{3}$ and $\sqrt{27}$ into $\sqrt{81}$, and that we can split out $\sqrt{18}$ into its components of $\sqrt{2}$ and $\sqrt{9}$.

$$\begin{gathered} 5\sqrt{3}\ast \sqrt{27}-2\sqrt{81}-\sqrt{18} \\ 5\sqrt{3\ast 27}-2\sqrt{81}-\sqrt{2\ast 9} \\ 5\sqrt{81}-2\sqrt{81}-\sqrt{2}\ast \sqrt{9} \\ 5\ast 9-2\ast 9-\sqrt{2}\ast 3 \\ 45-18-3\sqrt{2} \\ 27-3\sqrt{2} \end{gathered}$$

So, the equation simplifies to $27-3\sqrt{2}$.

Bringing it all together: question walkthrough video

Here’s a video going through one of our practice questions to demonstrate these ideas in action: