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Simplifying Square Roots

We now learn how to simplify square roots of numbers. 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.

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.

Tutorial 1


Exercise 1

Simplify each of the following:

  1. \(\sqrt{12}\)

  2. \(\sqrt{32}\)

  3. \(\sqrt{108}\)

  4. \(\sqrt{75}\)

  5. \(\sqrt{80}\)

  6. \(\sqrt{28}\)

  7. \(\sqrt{147}\)

Answers Without Working

  1. \(\sqrt{12} = 2\sqrt{3}\)

  2. \(\sqrt{32} = 3 \sqrt{2}\)

  3. \(\sqrt{108} = 6 \sqrt{3}\)

  4. \(\sqrt{75} = 5\sqrt{3}\)

  5. \(\sqrt{80} = 4\sqrt{5}\)

  6. \(\sqrt{28} = 2 \sqrt{7}\)

  7. \(\sqrt{147} = 7 \sqrt{3}\)

Solution With Working

  1. 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: \[\begin{aligned} \sqrt{12} & = \sqrt{4 \times 3} \\ & = \sqrt{4} \times \sqrt{3} \\ & = 2 \times \sqrt{3} \\ \sqrt{12} = 2\sqrt{3} \end{aligned}\]

  2. 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: \[\begin{aligned} \sqrt{32} & = \sqrt{16\times 2} \\ & = \sqrt{16}\times \sqrt{2} \\ & = 4 \times \sqrt{2} \\ \sqrt{32} &= 4\sqrt{2} \end{aligned} \]

  3. 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: \[\begin{aligned} \sqrt{108} & = \sqrt{36\times 3} \\ & = \sqrt{36} \times \sqrt{3} \\ & = 6 \times \sqrt{3} \\ \sqrt{108} & = 6 \sqrt{3} \end{aligned}\]

  4. 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: \[\begin{aligned} \sqrt{75} & = \sqrt{25 \times 3} \\ & = \sqrt{25} \times \sqrt{3} \\ & = 5 \times \sqrt{3} \\ \sqrt{75} & = 5\sqrt{3} \end{aligned}\]

  5. 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: \[\begin{aligned} \sqrt{80} & = \sqrt{16 \times 5} \\ & = \sqrt{16} \times \sqrt{5} \\ & = 4 \times \sqrt{5} \\ \sqrt{80} & = 4 \sqrt{5} \end{aligned}\]

  6. 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: \[\begin{aligned} \sqrt{28} & = \sqrt{4 \times 7} \\ & = \sqrt{4} \times \sqrt{7} \\ & = 2 \times \sqrt{7} \\ \sqrt{28} & = 2 \sqrt{7} \end{aligned}\]

  7. 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: \[\begin{aligned} \sqrt{147} & = \sqrt{49 \times 3} \\ & = \sqrt{49}\times \sqrt{3} \\ & = 7 \times \sqrt{3} \\ \sqrt{147} & = 7 \sqrt{3} \end{aligned}\]

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


Exercise 2

Simplify each of the following:

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

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

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

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

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

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

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

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

Solution Without Working

  1. \(\sqrt{\frac{25}{9}} = \frac{5}{3}\)

  2. \(\sqrt{\frac{4}{49}} = \frac{2}{7}\)

  3. \(\sqrt{\frac{72}{75}} = \frac{6\sqrt{2}}{5\sqrt{3}}\)

  4. \(\sqrt{\frac{27}{128}} = \frac{3\sqrt{3}}{8\sqrt{2}}\)

  5. \(\sqrt{\frac{48}{81}} = \frac{4\sqrt{3}}{9}\)

  6. \(\frac{\sqrt{32}}{\sqrt{2}} = 4 \)

  7. \(\frac{\sqrt{150}}{\sqrt{3}} = 5\sqrt{2}\)

  8. \(\frac{\sqrt{225}}{\sqrt{75}} = \sqrt{3}\)

Solution With Working

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

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

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

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

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

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

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

  8. 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

Exercise 3

Simplify each of the following as much as possible:

  1. \(2\sqrt{18} + 3 \sqrt{50}\)

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

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

  4. \(5\sqrt{12}-5\sqrt{72} + \sqrt{48}\)

  5. \(3\sqrt{\frac{8}{27}} + 6 \sqrt{\frac{32}{75}}\)

Solutions Without Working

  1. \(2\sqrt{18} + 3 \sqrt{50} = 21\sqrt{2}\)

  2. \(3\sqrt{5} - 2\sqrt{45} + 8 \sqrt{20} = 13\sqrt{5}\)

  3. \(\sqrt{200} - \sqrt{8} + 3 \sqrt{32} = 20\sqrt{2}\)

  4. \(5\sqrt{12} - 5\sqrt{72} + \sqrt{48} = 14\sqrt{3} - 30\sqrt{2}\)

  5. \(3\sqrt{\frac{8}{27}} + 6 \sqrt{\frac{32}{75}} = \frac{10\sqrt{2}}{\sqrt{3}}\)

Solution With Working

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

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

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

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

  5. 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: \[3\sqrt{\frac{8}{27}} + 6 \sqrt{\frac{32}{75}} = \frac{34}{5} \sqrt{\frac{2}{3}} \]