# Arron's Test wiki

**Simplifying radicals** is the process of manipulating a radical expression into a simpler or alternate form. Generally speaking, it is the process of simplifying expressions applied to radicals.

#### Contents

## Introduction

A radical is a number that has a fraction as its exponent.

\[ \sqrt[n]{x^m} = x ^ { m/n }. \]

Since they are exponents, radicals can be simplified using rules of exponents.

## Simplifying Simple Radicals

The square root of a positive non-perfect square is always an irrational number. The decimal representation of such a number loses precision when it is rounded, and it is time-consuming to compute without the aid of a calculator. Instead of using decimal representation, the standard way to write such a number is to use simplified radical form, which involves writing the radical with no perfect squares as factors of the number under the root symbol.

Let \(a\) be a positive non-perfect square integer.

The

simplified radical formof the square root of \(a\) is\[\sqrt{a}=b\sqrt{c}.\]

In this form \(\sqrt{a}=b\sqrt{c}\), both \(b\) and \(c\) are positive integers, and \(c\) contains no perfect square factors other than \(1\).

The process for putting a square root into simplified radical form involves finding perfect square factors, then applying the identity:\(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\), which allows us to take the root of the perfect square factors.

## Simplify \(\sqrt{12} \).

Since \( 12 = 2 \times 2 \times 3= 2^2 \times 3 \), we can rewrite it as \( \sqrt{12} = \sqrt{ 2^2 \times 3 } = \sqrt{2^2} \times \sqrt{3} = 2 \sqrt{3}.\)

Simplify \(\sqrt{72}\).

First, ask yourself, "What is a perfect square factor of \(72\)?"\(4\) is a perfect square factor of \(72\), and \(9\) is a perfect square factor of \(72\).

For the sake of this process, it is more efficient to find the

largestperfect square factor of \(72\). As shown below, \(4\times 9=36\) is thelargestperfect square factor of \(72\):\[\begin{align} \sqrt{72}&=\sqrt{36\times 2} \\ &=\sqrt{36}\times\sqrt{2} \\ &=6\sqrt{2}. \end{align}\]

Therefore, simplified form of \(\sqrt{72}\) is \(6\sqrt{2}.\ _\square\)

Note: When a number is placed to the left of a square root symbol, multiplication is implied. "\(6\sqrt{2}\)" is read as "\(6\) times the square root of \(2\)." \(_\square\)

Similarly, roots of higher degree (cube roots, fourth roots, etc.) are simplified when they have no factors under the radical that are perfect powers of the same degree as the radical.

## Simplify \(\sqrt[3] { a^2 b^ 4 } \).

Notice that we have \( b^3 \), which is a cube factor in the radicand. Hence, we can pull it out to obtain

\[ \sqrt[3] { a^2 b^4 } = b \sqrt[3] { a^2b}.\ _\square\]

## Adding Radicals

Since radicals are actually exponential expressions, they follow the rules of exponents and cannot be added together. In particular, you should avoid the common mistake shown below:

\[ \sqrt{a+b} \neq \sqrt{a}+\sqrt{b}.\]

This means that when we are dealing with radicals with different radicands, like \(sqrt{5}\) and \sqrt{7}), there is really no way to combine or simplify them. However, when dealing with radicals that share a base, we *can* simplify them by combining like terms.

## Simplify \(\sqrt{12} + 3\sqrt{3}\).

Since \(\sqrt{12} = 2\sqrt{3}\), we have \(\sqrt{12} + 3\sqrt{3}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}.\)

## Multiplying Radicals

When multiplying radicals, we make extensive use of the identity \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\). This means that two radicals, when multiplied together, might produce an integer rather than another radical.

## Simplify \(\sqrt{12} \times \sqrt{3}\).

Since \(12 \times 3 = 36\), we have \(\sqrt{12} \times \sqrt{3} = \sqrt{36} = 6\).

This will not always happen; however, the identity used above can be applied even to more complicated radical multiplication.

## Expand \( \left(1+\sqrt{21}\right)\left( 3-\sqrt{3}\right)\).

Using the distributive property and the identity \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\), we have\[\begin{align} \left(1+\sqrt{2}\right)\left( 3-\sqrt{3}\right) &= 1\left(3-\sqrt{3}\right) + \sqrt{21}\left(3-\sqrt{3}\right) \\ &= 3-\sqrt{3} + 3\sqrt{21} - \sqrt{21} \sqrt{3} \\ &= 3 - \sqrt{3} + 3\sqrt{21} - \sqrt{63} \\ &= 3 - \sqrt{3} + 3\sqrt{21} - 3\sqrt{7}. \end{align}\]

## Rationalizing Denominators

For more detail, refer to Rationalizing Denominators

Fractions are not considered to be written in simplest form if they have a irrational number (like \(\sqrt{2}\), for example) in the denominator. We can simplify the fraction by *rationalizing the denominator*. This is a procedure that frequently appears in problems involving radicals.

For problems involving simple radicals, the approach is fairly simple.

Simplify \(\frac{2}{\sqrt{3}}\).

We can simplify this fraction by multiplying by \(1=\frac{\sqrt{3}}{\sqrt{3}}\). Thus, we have \(\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}= \frac{2\sqrt{3}}{3}\).

When we have a more complicated expression involving a radical, we must look for another approach. However, when the denominator is a binomial expression involving radicals, we can use the difference of two squares identity to produce a conjugate pair that will remove the radicals from the denominator. For example, if we want to remove the radicals from the expression \(\sqrt{2}+\sqrt{3}\), we can multiply it by its conjugate pair \(\sqrt{2} - \sqrt{3}\).

Simplify \(\frac{1}{\sqrt{2}+\sqrt{3}}\).

\[\frac{1}{\sqrt{2}+\sqrt{3}} \cdot \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}} = \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2}^2 - \sqrt{6} + \sqrt{6} - \sqrt{3}^2} = \frac{\sqrt{2} - \sqrt{3}}{2-3} = \frac{\sqrt3-\sqrt2}{1} = \sqrt3 - \sqrt2\]

## See Also

**Cite as:**Arron's Test wiki.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/arrons-test-wiki/