# Simplifying Radicals

**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.

## Introduction

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

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

A square root \(\left( \sqrt{x} \right)\) is simplified, when the radicand (value inside the square root sign, which is \(x\) in this case) has no square factors. This can be done effectively by factoring the radicand first, to know what terms to pull out.

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

Since \( 12 = 2 \times 2 \times 3= 2^2 \times 3 \), \( \sqrt{12} = \sqrt{ 2^2 \times 3 } = 2 \sqrt{3} .\) \(_\square\)

Similarly, cube roots (or any root of degree \(n\)) are simplified when they have no cube (or degree \(n\)) factors.

## 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

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

Try the following problems.

\( \begin{eqnarray} \text{Statement 1: } \sqrt{x^2} &=& x \\\\ \text{Statement 2: } \sqrt{x^4} &=& x^2 \\\\ \text{Statement 3: } \sqrt{x^6} &=& x^3 \\\\ \text{Statement 4: } \sqrt{x^8} &=& x^4 \\\\ \end{eqnarray} \)

If \(x\) is a real number, which of the statements above are true?

## Useful Techniques

To simplify the expressions with radicals, one needs to know the rules of exponents.

Based on these rules, here are some examples.

## If you write the prime factorization of\( \sqrt[5]{32^8}\), what is the sum of the indices of the factors?

If you recognize that \( 32 = 2^5 \), the answer falls quickly into place as shown below:

\[ \sqrt[5]{32^8} = \left(2^5\right)^\frac{8}{5} = 2^8 .\]

So, the sum of the indices is simply \(8.\) \( _\square \)

## \(\dfrac{\sqrt[10]{{1024}^{100}}}{\sqrt[100]{{1024}^{10}}}\) can be expressed as \(2^a\). Find \(a\).

\[\begin{align} \dfrac{\sqrt[10]{{1024}^{100}}}{\sqrt[100]{{1024}^{10}}} & = \dfrac{\sqrt[10]{(2^{10})^{100}}}{\sqrt[100]{(2^{10})^{10}}}\\\\ & = \dfrac {\left((2^{10})^{100}\right)^{\frac {1}{10}}}{\left((2^{10})^{10}\right)^{\frac{1}{100}}}\\\\ & = \dfrac {2^{\frac {1}{10} × 100 × 10}}{2^{\frac {1}{100} × 10 × 10}}\\\\ & = \dfrac {2^{100}}{2}\\\\ & = 2^{(100 - 1)}\\\\ & = 2^{99}\\\\ \Rightarrow a & = 99.\ _\square \end {align}\]

Try the following problems at your own.

One of the most useful techniques for simplifying the expressions with radicals is rationalizing denominators.

Consider the fraction with irrational denominator \(\frac{a}{\sqrt{b}}\). Rationalizing the denominator, we get the above expression as \[\begin{align} \frac{a}{\sqrt{b}} &=\frac{a}{\sqrt{b} } \times 1\\ &=\frac{a}{\sqrt{b} } \times \frac{\sqrt{b}}{\sqrt{b}} \\ &=\frac{a\sqrt{b}}{b}. \end{align}\]

Here are the examples worked out on the basis of this technique.

Simplify \( \displaystyle \frac{1}{2+\sqrt{5}} \).

To ease in the calculation, we must remove the irrationals from the denominator and bring it in the numerator. We have

\[\begin{align} \frac{1}{2+\sqrt{5}} &= \frac{1}{2+\sqrt{5}} \left( \frac{2-\sqrt{5}}{2-\sqrt{5}} \right) \\ &= \frac{2-\sqrt{5}}{2^2 -5} \\ &= -2 + \sqrt{5}. \ _\square \end{align} \]

Simplify \(\dfrac{3\sqrt 5 - 4\sqrt 2}{2\sqrt 5 - 3\sqrt 2}\).

To make the expression simpler, we multiply and divide it with the conjugate of the denominator as follows:

\[\begin{align} \dfrac{3\sqrt 5 - 4\sqrt 2}{2\sqrt 5 - 3\sqrt 2} &= \dfrac{3\sqrt 5 - 4\sqrt 2}{2\sqrt 5 - 3\sqrt 2} \times \dfrac{2\sqrt 5 + 3\sqrt 2}{2\sqrt 5 + 3\sqrt 2}\\\\ &=\dfrac{30 - 8\sqrt{10} + 9\sqrt{10} - 24}{\big(2\sqrt 5\big)^2-\big(3\sqrt 2\big)^2}\\\\ &=\dfrac{6+\sqrt{10}}{20-18}\\\\ &=\dfrac{6+\sqrt{10}}{2}.\ _\square \end{align}\]

Simplify \(\dfrac {11 - \sqrt {10}}{11+ \sqrt{10}}\).

We have

\[\begin{align} \dfrac {11 - \sqrt {10}}{11+ \sqrt{10}} & = \dfrac {(11 - \sqrt {10})\big(11 - \sqrt{10}\big)}{\big(11+ \sqrt{11}\big)\big(11 - \sqrt {10}\big)}\\\\ & = \dfrac {{(11)}^2 + {\big(\sqrt {10}\big)}^2 - 2(11)\big(\sqrt {10}\big)} {{(11)}^2 - {\big(\sqrt {10}\big)}^2}\\\\ & = \dfrac {121 + 10 - 22\sqrt {10}}{121 - 10}\\\\ & = \dfrac {131 - 22\sqrt {10}}{111}.\ _\square \end {align}\]

Try the following problems for practicing the technique of rationalizing denominators:

## Problem Solving

This section contains the examples and the problems that will make you better at simplifying expressions with radicals.

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

Note that \( 3 + 2 \sqrt{2} = 1^{2} + \left(\sqrt{2}\right)^{2} + 2 \times 1 \times \sqrt{2} = \left(1 + \sqrt{2} \right) ^{2}. \)

Hence, \( \sqrt{3 + 2 \sqrt{2} } = \sqrt{\left(1 + \sqrt{2} \right) ^{2}} = \left| 1 + \sqrt{2} \right| = 1 + \sqrt{2} . \ _\square \)

Simplify \( \large \sqrt{\frac{x+\sqrt{x^2-y}}{2}} + \sqrt{\frac{x-\sqrt{x^2-y}}{2}} \).

Let's call this identity \( A \) and \( z = \sqrt{x^2 - y} \). Then \[ A^2 = \frac{x + z}{2} + \frac{x - z}{2} + 2 \sqrt{ \frac{x + z}{2} \cdot \frac{x - z}{2}}, \] and simplyfing \( (x+z)(x-z),\) we have \( A^2 = x + \sqrt{y} \). Thus \( A = \sqrt{x + \sqrt{y}} \). \(_\square\)

Simplify \(\dfrac { 1 }{ \sqrt [ 3 ]{ 3 } -\sqrt [ 3 ]{ 2 } } \).

We know we can simplify radicals and irrationals using simple algebraic identities. In this we will use

\[a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+ab\right).\]

Now we can solve it as follows:

\[\begin{align} \dfrac { { \left( \sqrt [ 3 ]{ 3 } \right) }^{ 3 }-{ \left( \sqrt [ 3 ]{ 2 } \right) }^{ 3 } }{ \sqrt [ 3 ]{ 3 } -\sqrt [ 3 ]{ 2 } } &=\dfrac { \left( \sqrt [ 3 ]{ 3 } -\sqrt [ 3 ]{ 2 } \right) \left( \sqrt [ 3 ]{ { 3 }^{ 2 } } +\sqrt [ 3 ]{ { 2 }^{ 2 } } +\sqrt [ 3 ]{ 3\times 2 } \right) }{ \sqrt [ 3 ]{ 3 } -\sqrt [ 3 ]{ 2 } } \\ &=\sqrt [ 3 ]{ 4 } +\sqrt [ 3 ]{ 6 } +\sqrt [ 3 ]{ 9 }.\ _\square \end{align} \]

Simplify \(\dfrac{\big({\sqrt{5}\big)}^3 - {\big(\sqrt{4}\big)}^3}{{\big(\sqrt{5}\big)}^3 + {\big(\sqrt{4}\big)}^3}\).

We have

\[\begin {align} \dfrac{{\big(\sqrt{5}\big)}^3 - {\big(\sqrt{4}\big)}^3}{{\big(\sqrt{5}\big)}^3 + {\big(\sqrt{4}\big)}^3} & = \dfrac{\big(\sqrt 5 - \sqrt 4\big)\big(5 + \sqrt{20} + 4\big)}{\big(\sqrt 5 + \sqrt 4\big)\big(5 - \sqrt{20} + 4\big)}\\ & = \dfrac{\big(\sqrt{5} - \sqrt{4}\big)\big(9 + \sqrt{5}\big)}{\big(\sqrt{5} + \sqrt{4}\big)\big(9 - \sqrt{5}\big)}\\ & = \dfrac{\big(\sqrt{5} - \sqrt{4}\big)\big(9 + \sqrt{5}\big)\big(9 + \sqrt{5}\big)}{\big(\sqrt{5} + \sqrt{4}\big)\big(9 - \sqrt{5}\big)\big(9+\sqrt{5}\big)}\\ & = \dfrac{\big(\sqrt{5} - 2\big)\big(81 + 5 + 18\sqrt{5}\big)}{\big(\sqrt{5} + 2\big)(81-5)}\\ & = \dfrac{\big(\sqrt 5 - 2\big)\big(86 + 18\sqrt {5}\big)}{\big(\sqrt 5 + 2\big)(84)}\\ & = \dfrac{\big(86 + 18\sqrt {5}\big)\big(\sqrt 5 - 2\big)}{(84)\big(\sqrt 5 + 2\big)}\\ & = \dfrac{\big(86 + 18\sqrt {5}\big)\big(\sqrt 5 - 2\big)\big(\sqrt 5 - 2\big)}{(84)\big(\sqrt 5 + 2\big)\big(\sqrt 5 - 2\big)}\\ & = \dfrac{\big(86 + 18\sqrt {5}\big)\big(5 + 4 - 4\sqrt 5\big)}{(84)(5-4)}\\ & = \dfrac{\big(86 + 18\sqrt {5}\big)\big(9 - 4\sqrt 5\big)}{84}\\ & = \dfrac {414 - 182\sqrt 5}{84}\\ & = \dfrac {212 - 91\sqrt 5}{42}.\ _\square \end {align}\]

Try the following problems:

## See Also

**Cite as:**Simplifying Radicals.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/simplify-radicals/