SIMPLIFYING SQUARE ROOTS AND RADICAL FRACTIONS

There are times when we need to simplify a square root such as √500 into a simpler form, that is 10√5. Similarly, the form $\sqrt{\frac{5}{3}}$ can be simplified to $\frac{1}{3} \sqrt{15}$ . Additionally, the radical fraction $\frac{2}{\sqrt{5}+\sqrt{3}}$ can be simplified to (√5 – √3). This post discusses how to convert such forms into simpler ones.

Simplifying √a if some factor of a is a perfect square; a ≥ 0

Express a as the product of factors some of which are perfect squares. Then, pull the roots of the perfect squares out of the radical sign. Suppose that a = a₁²⋅b where b does not have any perfect square (other than 1) as its factor. Then √a = a₁√b.

Example 1

Simplify √500.

Answer

500 can be expressed as the product of factors some of which is a perfect square. It is the product of 100 (a perfect square) and 5. In short, 500 = 10²⋅5. Pulling the root of 10² out of the radical sign results in:

$\sqrt{500}=\sqrt{10^2 \cdot 5}=10 \sqrt{5}$

Thus, √500 = 10√5.

Example 2

Simplify √216.

Answer

216 can be expressed as the product of factors some of which is a perfect square. It is the product of 36 (a perfect square) and 6, i.e. 216 = 6²⋅6. Pulling the root of 6² out of the radical sign yields:

$\sqrt{216}=\sqrt{6^2 \cdot 6}=6 \sqrt{6}$

So, √216 = 6√6.

Simplifying the form of $\sqrt{\frac{a}{b}}$ ; ab > 0

Multiply the numerator and denominator of $\frac{a}{b}$ by b so that the denominator of the fraction under the radical sign is a perfect square, i.e. b². We can pull the root of the perfect square out of the radical sign as follows.

$\sqrt{\frac{a}{b}}=\sqrt{\frac{ab}{b^2}}=\frac{1}{b} \sqrt{ab}$

Example 3

Simplify $\sqrt{\frac{5}{3}}$ .

Answer

Multiply the numerator and denominator of $\frac{5}{3}$ by 3 so that the denominator of the fraction under the radical sign is a perfect square, i.e. 3². Pull the root of the perfect square out of the radical sign as follows.

$\sqrt{\frac{5}{3}}=\sqrt{\frac{15}{9}}=\frac{1}{3} \sqrt{15}$ .

Therefore, $\sqrt{\frac{5}{3}}=\frac{1}{3} \sqrt{15}$ .

Example 4

Simplify $\sqrt{\frac{10}{7}}$ .

Answer

Multiply the numerator and denominator of $\frac{10}{7}$ by 7 so that the denominator of the fraction under the radical sign is a perfect square, i.e. 7². Pull the root of the perfect square out of the radical sign as follows.

$\sqrt{\frac{10}{7}}=\sqrt{\frac{70}{49}}=\frac{1}{7} \sqrt{70}$ .

In conclusion, $\sqrt{\frac{10}{7}}=\frac{1}{7} \sqrt{70}$ .

Simplifying the form of $\frac{a}{\sqrt{b}}$ ; b > 0

Multiply the numerator and denominator of $\frac{a}{\sqrt{b}}$ by √b. This results in:

$\frac{a}{\sqrt{b}}=\frac{a}{\sqrt{b}} \cdot \frac{\sqrt{b}}{\sqrt{b}}=\frac{ a}{b} \sqrt{b}$

Example 5

Simplify $\frac{5}{\sqrt{11}}$ .

Answer

Multiply the numerator and denominator of $\frac{5}{\sqrt{11}}$ by √11. It follows that:

$\frac{5}{\sqrt{11}}=\frac{5}{\sqrt{11}} \cdot \frac{\sqrt{11}}{\sqrt{11}}=\frac{ 5}{11} \sqrt{11}$

Example 6

Simplify $\frac{7-\sqrt{5}}{\sqrt{5}}$ .

Answer

Multiply the numerator and denominator of $\frac{7-\sqrt{5}}{\sqrt{5}}$ by √5 to get:

$\frac{7-\sqrt{5}}{\sqrt{5}}=\frac{7-\sqrt{5}}{\sqrt{5}} \cdot \frac{\sqrt{5} }{\sqrt{5}}$

$\frac{7-\sqrt{5}}{\sqrt{5}}=\frac{(7-\sqrt{5}) \sqrt{5}}{5}$

$\frac{7-\sqrt{5}}{\sqrt{5}}=\frac{7 \sqrt{5}-5}{5}$

Simplifying the form of $\frac{a}{\sqrt{b} + \sqrt{c}}$ or $\frac{a}{\sqrt{b} - \sqrt{c}}$

To simplify a radical fraction of such forms, we multiply the numerator and denominator by the conjugate of the denominator. The conjugate of (√b + √c) is (√b – √c), whereas the conjugate of (√b – √c) is (√b + √c). As a consequence, to simplify the form $\frac{a}{\sqrt{b} + \sqrt{c}}$ , we multiply both the numerator and denominator by the conjugate of (√b + √c ), i.e. (√b – √c). The technique is demonstrated as follows.

$\frac{a}{\sqrt{b} + \sqrt{c}} = \frac{a}{\sqrt{b} + \sqrt{c}} \cdot \frac{\sqrt{b} - \sqrt{c}}{\sqrt{b} - \sqrt{c}}$

$\frac{a}{\sqrt{b} + \sqrt{c}} = \frac{a(\sqrt{b} - \sqrt{c})}{(\sqrt{b})^2 - (\sqrt{c})^2}$

$\frac{a}{\sqrt{b} + \sqrt{c}} = \frac{a(\sqrt{b} - \sqrt{c})}{b - c}$

Example 7

Simplify $\frac{3}{\sqrt{7} + \sqrt{6}}$ .

Answer

The conjugate of (√7 + √6) is (√7 – √6). To simplify the radical fraction, multiply it by $\frac{\sqrt{7} - \sqrt{6}}{\sqrt{7} - \sqrt{6}}$ . It follows that:

$\frac{3}{\sqrt{7} + \sqrt{6}} = \frac{3}{\sqrt{7} + \sqrt{6}} \cdot \frac{\sqrt{7} - \sqrt{6}}{\sqrt{7} - \sqrt{6}}$

$\frac{3}{\sqrt{7} + \sqrt{6}} = \frac{3(\sqrt{7} - \sqrt{6})}{(\sqrt{7})^2 - (\sqrt{6})^2}$

$\frac{3}{\sqrt{7} + \sqrt{6}} = \frac{3(\sqrt{7} - \sqrt{6})}{7 - 6}$

$\frac{3}{\sqrt{7} + \sqrt{6}} = 3 \sqrt{7} - 3 \sqrt{6}$

Example 8

Simplify $\frac{\sqrt{5}}{\sqrt{5} + \sqrt{2}}$ .

Answer

The conjugate of (√5 + √2) is (√5 – √2). To simplify the fraction, multiply it by $\frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}}$ . It follows that:

$\frac{\sqrt{5}}{\sqrt{5} + \sqrt{2}} = \frac{\sqrt{5}}{\sqrt{5} + \sqrt{2}} \cdot \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}}$

$\frac{\sqrt{5}}{\sqrt{5} + \sqrt{2}} = \frac{{\sqrt{5}}(\sqrt{5} - \sqrt{2})} {(\sqrt{5})^2 - (\sqrt{2})^2}$

$\frac{\sqrt{5}}{\sqrt{5} + \sqrt{2}} = \frac{5}{3} - \frac{1}{3} \sqrt{10}$

To simplify the form $\frac{a}{\sqrt{b} - \sqrt{c}}$ , multiply both the numerator and denominator by the conjugate of (√b – √c ), that is (√b + √c). Hence:

$\frac{a}{\sqrt{b} - \sqrt{c}} = \frac{a}{\sqrt{b} - \sqrt{c}} \cdot \frac{\sqrt{b} + \sqrt{c}}{\sqrt{b} + \sqrt{c}}$

$\frac{a}{\sqrt{b} - \sqrt{c}} = \frac{a(\sqrt{b} + \sqrt{c})}{(\sqrt{b})^2 - (\sqrt{c})^2}$

$\frac{a}{\sqrt{b} - \sqrt{c}} = \frac{a(\sqrt{b} + \sqrt{c})}{b - c}$

Example 9

Simplify $\frac{3}{\sqrt{7} - \sqrt{6}}$ .

Answer

The conjugate of (√7 – √6) is (√7 + √6). To simplify the fraction, multiply it by $\frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}}$ . This results in: