How to Simplify Square Roots and Surds

We now learn how to simplify surds, that's how to write a square root in its simplest form.

Our objective, by the end of this section will be to know how to simplify radical expresssions looking like either of the following:

\(\sqrt{108}\), or \(\sqrt{\frac{72}{75}}\), or even \(2\sqrt{18}+3\sqrt{50} - 5 \sqrt{12}\).

We start by learning how to simplify a square root, by looking for square number factors of the radicand (remember the radicand is the number inside the square root) as well as by using the fact that the square root of a product of two numbers equals the product of the square roots of those numbers, \(\sqrt{a\times b} = \sqrt{a} \times \sqrt{b}\).
Tutorials and exercises with solutions will help us along the way.

Tutorial : How to Simplify Square Roots

In this video we learn how to simplify square roots.

Method - Simplifying Roots

Given a square root, we can simplify it using the following three steps:

Step 1: write a list of the first few square numbers: \[1,\ 4, \ 9, \ 16, \ 25, \ 36, \ 49, \ \dots \]

Step 2: look for largest factor of the radicand in the list of square numbers, from step 1.

Step 3: use the fact that: \[\sqrt{a\times b} = \sqrt{a}\times \sqrt{b}\] to simplify the root.

Exercise 1

Simplify each of the following:

\(\sqrt{8}\)

\(\sqrt{12}\)

\(\sqrt{32}\)

\(\sqrt{108}\)

\(\sqrt{75}\)

\(\sqrt{80}\)

\(\sqrt{28}\)

\(\sqrt{147}\)

Solution

To simplify \(\sqrt{12}\) we follow our three steps:
- Step 1: list the first few square numbers: \[1,4,9,16,25, \dots \]
- Step 2: look for the largest factor of \(8\) from the list, written in step 1.
  
  The largest factor is \(4\), indeed: \[8 = 4\times 2\]
- Step 3: We now use the fact that \(\sqrt{a\times b} = \sqrt{a}\times \sqrt{b}\) to simplify the square root: \[\begin{aligned} \sqrt{8} & = \sqrt{4 \times 2} \\ & = \sqrt{4} \times \sqrt{2} \\ & = 2 \times \sqrt{2} \\ \sqrt{12} = 2\sqrt{2} \end{aligned}\]

To simplify \(\sqrt{12}\) we follow our three steps:
- Step 1: list the first few square numbers: \[1,4,9,16,25, \dots \]
- Step 2: look for the largest factor of \(12\) from the list, written in step 1.
  
  The largest factor is \(4\), indeed: \[12 = 4\times 3\]
- Step 3: We now use the fact that \(\sqrt{a\times b} = \sqrt{a}\times \sqrt{b}\) to simplify the square root: \[\begin{aligned} \sqrt{12} & = \sqrt{4 \times 3} \\ & = \sqrt{4} \times \sqrt{3} \\ & = 2 \times \sqrt{3} \\ \sqrt{12} = 2\sqrt{3} \end{aligned}\]

We simplify \(\sqrt{32}\) as follows:
- Step 1: list the first few square numbers: \[1,4,9,16,25, \dots \]
- Step 2: look for the largest factor of \(32\) from the list, written in step 1.
  
  The largest factor is \(16\), indeed: \[32 = 16\times 2\]
- Step 3: We now use the fact that \(\sqrt{a\times b} = \sqrt{a}\times \sqrt{b}\) to simplify the square root: \[\begin{aligned} \sqrt{32} & = \sqrt{16\times 2} \\ & = \sqrt{16}\times \sqrt{2} \\ & = 4 \times \sqrt{2} \\ \sqrt{32} &= 4\sqrt{2} \end{aligned} \]

We simplify \(\sqrt{108}\) as follows:
- Step 1: list the first few square numbers: \[1,4,9,16,25, 36, 49, 64, \dots \]
- Step 2: look for the largest factor of \(108\) from the list, written in step 1.
  
  The largest factor is \(36\), indeed: \[108 = 36\times 3\]
- Step 3: We now use the fact that \(\sqrt{a\times b} = \sqrt{a}\times \sqrt{b}\) to simplify the square root: \[\begin{aligned} \sqrt{108} & = \sqrt{36\times 3} \\ & = \sqrt{36} \times \sqrt{3} \\ & = 6 \times \sqrt{3} \\ \sqrt{108} & = 6 \sqrt{3} \end{aligned}\]

We simplify \(\sqrt{75}\) as follows:
- Step 1: list the first few square numbers: \[1,4,9,16,25, 36, 49, \dots \]
- Step 2: look for the largest factor of \(75\) from the list, written in step 1.
  
  The largest factor is \(25\), indeed: \[75 = 25\times 3\]
- Step 3: We now use the fact that \(\sqrt{a\times b} = \sqrt{a}\times \sqrt{b}\) to simplify the square root: \[\begin{aligned} \sqrt{75} & = \sqrt{25 \times 3} \\ & = \sqrt{25} \times \sqrt{3} \\ & = 5 \times \sqrt{3} \\ \sqrt{75} & = 5\sqrt{3} \end{aligned}\]

We simplify \(\sqrt{80}\) as follows:
- Step 1: list the first few square numbers: \[1,4,9,16,25, 36, 49, \dots \]
- Step 2: look for the largest factor of \(80\) from the list, written in step 1.
  
  The largest factor is \(16\), indeed: \[80 = 16\times 5\]
- Step 3: We now use the fact that \(\sqrt{a\times b} = \sqrt{a}\times \sqrt{b}\) to simplify the square root: \[\begin{aligned} \sqrt{80} & = \sqrt{16 \times 5} \\ & = \sqrt{16} \times \sqrt{5} \\ & = 4 \times \sqrt{5} \\ \sqrt{80} & = 4 \sqrt{5} \end{aligned}\]

We simplify \(\sqrt{28}\) as follows:
- Step 1: list the first few square numbers: \[1,4,9,16,25, \dots \]
- Step 2: look for the largest factor of \(28\) from the list, written in step 1.
  
  The largest factor is \(4\), indeed: \[28 = 4\times 7\]
- Step 3: We now use the fact that \(\sqrt{a\times b} = \sqrt{a}\times \sqrt{b}\) to simplify the square root: \[\begin{aligned} \sqrt{28} & = \sqrt{4 \times 7} \\ & = \sqrt{4} \times \sqrt{7} \\ & = 2 \times \sqrt{7} \\ \sqrt{28} & = 2 \sqrt{7} \end{aligned}\]

We simplify \(\sqrt{147}\) as follows:
- Step 1: list the first few square numbers: \[1,4,9,16,25, 36, 49, \dots \]
- Step 2: look for the largest factor of \(147\) from the list, written in step 1.
  
  The largest factor is \(49\), indeed: \[147 = 49 \times 3\]
- Step 3: We now use the fact that \(\sqrt{a\times b} = \sqrt{a}\times \sqrt{b}\) to simplify the square root: \[\begin{aligned} \sqrt{147} & = \sqrt{49 \times 3} \\ & = \sqrt{49}\times \sqrt{3} \\ & = 7 \times \sqrt{3} \\ \sqrt{147} & = 7 \sqrt{3} \end{aligned}\]

Solution With Working

Method - Simplifying Square Roots of Fractions

At times we may be faced with expressions involving fractions under a square root.
To deal with such expressions the rule that should always be kept in mind is: \[\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\] For instance we may need to simplify: \[\sqrt{\frac{16}{9}}\] The trick, therefore, is to write this as: \[\frac{\sqrt{16}}{\sqrt{9}} = \frac{4}{3} \] This is further explained in tutorial 2, below, watch it now to learn exactly how to deal with such square roots.

Tutorial 2: Fractions

In this video we learn how to simplify square roots that are expressed as fractions.

Exercise 2

Simplify each of the following:

\(\sqrt{\frac{25}{9}}\)

\(\sqrt{\frac{4}{49}}\)

\(\sqrt{\frac{72}{75}}\)

\(\sqrt{\frac{27}{128}}\)

\(\sqrt{\frac{48}{81}}\)

\(\frac{\sqrt{32}}{\sqrt{2}}\)

\(\frac{\sqrt{150}}{\sqrt{3}}\)

\(\frac{\sqrt{225}}{\sqrt{75}}\)

Solution

We simplify \(\sqrt{\frac{25}{9}}\) as follows: \[\begin{aligned} \sqrt{\frac{25}{9}} & = \frac{\sqrt{25}}{\sqrt{9}} \\ & = \frac{5}{3} \end{aligned}\]

We simplify \(\sqrt{\frac{4}{49}}\) as follows: \[\begin{aligned} \sqrt{\frac{4}{49}} & = \frac{\sqrt{4}}{\sqrt{49}} \\ & = \frac{2}{7} \end{aligned}\]

We simplify \(\sqrt{\frac{72}{75}}\) as follows: \[\begin{aligned} \sqrt{\frac{72}{75}} & = \frac{\sqrt{72}}{\sqrt{75}} \\ & = \frac{\sqrt{36\times 2}}{\sqrt{25\times 3}} \\ & = \frac{\sqrt{36}\times \sqrt{2}}{\sqrt{25} \times \sqrt{3}} \\ \sqrt{\frac{72}{75}} & = \frac{6 \sqrt{2}}{5\sqrt{3}} \end{aligned}\]

We simplify \(\sqrt{\frac{27}{128}}\) as follows: \[\begin{aligned} \sqrt{\frac{27}{128}} & = \frac{\sqrt{27}}{\sqrt{128}} \\ & = \frac{\sqrt{9\times 3}}{\sqrt{64\times 2}} \\ & = \frac{\sqrt{9}\times \sqrt{3}}{\sqrt{64}\times \sqrt{2}} \\ \sqrt{\frac{27}{128}} & = \frac{3\sqrt{3}}{8\sqrt{2}} \end{aligned}\]

We simplify \(\sqrt{\frac{48}{81}}\) as follows: \[\begin{aligned} \sqrt{\frac{48}{81}} &= \frac{\sqrt{48}}{\sqrt{81}} \\ & = \frac{\sqrt{16\times 3}}{9} \\ & = \frac{\sqrt{16}\times \sqrt{3}}{9}\\ \sqrt{\frac{48}{81}} & = \frac{4\sqrt{3}}{9} \end{aligned}\]

We simplify \(\frac{\sqrt{32}}{\sqrt{2}}\) as follows: \[\begin{aligned} \frac{\sqrt{32}}{\sqrt{2}} & = \sqrt{\frac{32}{2}} \\ & = \sqrt{16} \\ \frac{\sqrt{32}}{\sqrt{2}} & = 4 \end{aligned}\]

We simplify \(\frac{\sqrt{150}}{\sqrt{3}}\) as follows: \[\begin{aligned} \frac{\sqrt{150}}{\sqrt{3}} & = \sqrt{\frac{150}{3}} \\ & = \sqrt{50} \\ & = \sqrt{25\times 2}\\ & = \sqrt{25} \times \sqrt{2} \\ \frac{\sqrt{150}}{\sqrt{3}} & = 5\sqrt{2} \end{aligned}\]

We simplify \(\frac{\sqrt{225}}{\sqrt{75}}\) as follows: \[\begin{aligned} \frac{\sqrt{225}}{\sqrt{75}} & = \sqrt{\frac{225}{75}} \\ & = \sqrt{3} \end{aligned}\]

Simplifying Entire Expressions

We'll often be required to simplify entire expressions such as: \[2\sqrt{18} + 3\sqrt{50} - 5 \sqrt{12}\] To simplify such expressions we treat each radical (square root) as though it was on its own, using the techniques we learnt above.

We must also keep the following important rules in mind: \[\sqrt{a}+\sqrt{b} \neq \sqrt{a+b}\] We can only gather roots if the radicands (the number inside the root) are the same. For example: \[2\sqrt{3}+5\sqrt{3} = 7\sqrt{3}\] But \(2\sqrt{3}+5\sqrt{5}\) cannot be siimplified any further.

Watch tutorial 3 to see a few examples, then work through exercise 3.

Tutorial 3: Simplifying Entire Expressions

In this video we learn how to simplify entire expressions invovling square roots.

Exercise 3

Simplify each of the following as much as possible:

\(2\sqrt{18} + 3 \sqrt{50}\)

\(3\sqrt{5}-2\sqrt{45}+8\sqrt{20}\)

\(\sqrt{200} - \sqrt{8} + 3\sqrt{32}\)

\(5\sqrt{12}-5\sqrt{72} + \sqrt{48}\)

\(3\sqrt{\frac{8}{27}} + 6 \sqrt{\frac{32}{75}}\)

Solution

We simplify \(2\sqrt{18}+3\sqrt{50}\) as follows: \[\begin{aligned} 2\sqrt{18}+3\sqrt{50} & = 2\sqrt{9\times 2} + 3 \sqrt{25\times 2} \\ & = 2\sqrt{9}\times \sqrt{2} + 3\sqrt{25}\times \sqrt{2} \\ & = 2\times 3 \times \sqrt{2} + 3 \times 5 \times \sqrt{2} \\ & = 6\sqrt{2} + 15 \sqrt{2} \\ 2\sqrt{18}+3\sqrt{50} & = 21\sqrt{2} \end{aligned}\]

We simplify \(3\sqrt{5} - 2\sqrt{45}+ 8 \sqrt{20}\) as follows: \[\begin{aligned} 3\sqrt{5} - 2\sqrt{45}+ 8 \sqrt{20} & = 3\sqrt{5} - 2\sqrt{9\times 5} + 8 \sqrt{4\times 5} \\ & = 3\sqrt{5} - 2\times \sqrt{9} \times \sqrt{5} + 8 \times \sqrt{4} \times \sqrt{5} \\ & = 3\sqrt{5} - 2\times 3 \times \sqrt{5} + 8 \times 2 \times \sqrt{5} \\ & = 3\sqrt{5} - 6 \times \sqrt{5} + 16 \times \sqrt{5} \\ & = 3\sqrt{5} - 6 \sqrt{5} + 16 \sqrt{5} \\ & = -3\sqrt{5} + 16 \sqrt{5} \\ 3\sqrt{5} - 2\sqrt{45}+ 8 \sqrt{20} & = 13 \sqrt{5} \end{aligned}\]

We simplify \(\sqrt{200} - \sqrt{8} + 3\sqrt{32}\) as follows: \[\begin{aligned} \sqrt{200} - \sqrt{8} + 3\sqrt{32} & = \sqrt{100\times 2} - \sqrt{4\times 2} + 3\sqrt{16\times 2} \\ & = \sqrt{100} \times \sqrt{2} - \sqrt{4} \times \sqrt{2} + 3\times \sqrt{16} \times \sqrt{2} \\ & = 10 \times \sqrt{2} - 2 \times \sqrt{2} + 3\times 4 \times \sqrt{2} \\ & = 10\sqrt{2} - 2\sqrt{2} + 12\sqrt{2} \\ & = 8\sqrt{2} + 12\sqrt{2} \\ \sqrt{200} - \sqrt{8} + 3\sqrt{32} & = 20\sqrt{2} \end{aligned}\]

We simplify \(5\sqrt{12} - 5 \sqrt{72} + \sqrt{48}\) as follows: \[\begin{aligned} 5\sqrt{12} - 5 \sqrt{72} + \sqrt{48} & = 5\sqrt{4\times 3} - 5\sqrt{36\times 2} + \sqrt{16\times 3} \\ & = 5\times \sqrt{4} \times \sqrt{ 3} - 5\times \sqrt{36} \times \sqrt{2} + \sqrt{16} \times \sqrt{3} \\ & = 5\times 2 \times \sqrt{ 3} - 5\times 6 \times \sqrt{2} + 4 \times \sqrt{3} \\ & = 10 \times \sqrt{ 3} - 30 \times \sqrt{2} + 4 \times \sqrt{3} \\ & = 10\sqrt{ 3} - 30 \sqrt{2} + 4 \sqrt{3} \\ & = 10\sqrt{ 3} + 4 \sqrt{3} - 30 \sqrt{2} \\ 5\sqrt{12} - 5 \sqrt{72} + \sqrt{48} & = 14\sqrt{3} - 30\sqrt{2} \end{aligned}\]

We simplify \(3\sqrt{\frac{8}{27}} + 6 \sqrt{\frac{32}{75}}\) as follows: \[\begin{aligned} 3\sqrt{\frac{8}{27}} + 6 \sqrt{\frac{32}{75}} & = 3 \times \frac{\sqrt{8}}{\sqrt{27}} + 6 \times \frac{\sqrt{32}}{\sqrt{75}} \\ & = 3 \times \frac{\sqrt{4\times 2}}{\sqrt{9\times 3}} + 6 \times \frac{\sqrt{16\times 2}}{\sqrt{25 \times 3}} \\ & = 3 \times \frac{\sqrt{4} \times \sqrt{2}}{\sqrt{9} \times \sqrt{3}} + 6 \times \frac{\sqrt{16} \times \sqrt{2}}{\sqrt{25} \times \sqrt{3}} \\ & = 3 \times \frac{2\sqrt{2}}{3\sqrt{3}} + 6 \times \frac{4\sqrt{2}}{5\sqrt{3}} \\ & = \frac{3\times 2\sqrt{2}}{3\sqrt{3}} + \frac{6 \times 4\sqrt{2}}{5\sqrt{3}} \\ & = \frac{2\sqrt{2}}{\sqrt{3}} + \frac{24\sqrt{2}}{5\sqrt{3}} \\ & = \frac{5}{5} \times \frac{2\sqrt{2}}{\sqrt{3}} + \frac{24\sqrt{2}}{5\sqrt{3}} \\ & = \frac{10 \sqrt{2} + 24\sqrt{2}}{5\sqrt{3}} \\ 3\sqrt{\frac{8}{27}} + 6 \sqrt{\frac{32}{75}} & = \frac{34\sqrt{2}}{5\sqrt{3}} \end{aligned}\] We can also write the final answer as: \[3\sqrt{\frac{8}{27}} + 6 \sqrt{\frac{32}{75}} = \frac{34}{5} \sqrt{\frac{2}{3}} \]

Solution With Working

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RadfordMathematics.com

Online Mathematics Book

How to Simplify Square Roots and Surds

Tutorial : How to Simplify Square Roots

Method - Simplifying Roots

Exercise 1

Solution With Working

Method - Simplifying Square Roots of Fractions

Tutorial 2: Fractions

Exercise 2

Simplifying Entire Expressions

Tutorial 3: Simplifying Entire Expressions

Exercise 3

Solution With Working

RadfordMathematics.com

Online Mathematics Book

How to Simplify Square Roots and Surds

Tutorial : How to Simplify Square Roots

Method - Simplifying Roots

Exercise 1

Exercise 1 - Answer Key

Solution

Exercise 1 - Worked Solution

Solution

Solution With Working

Method - Simplifying Square Roots of Fractions

Tutorial 2: Fractions

Exercise 2

Exercise 2 - Answer Key

Solution

Exercise 2 - Worked Solution

Solution

Simplifying Entire Expressions

Tutorial 3: Simplifying Entire Expressions

Exercise 3

Exercise 2 - Answer Key

Solution

Exercise 3 - Worked Solution

Solution

Solution With Working