Rationalizing the denominator of a radical expression

Example #1:

 $$Rationalize \ \frac{3} {\sqrt{5}}$$
 $$Multiply \ by \ \frac{\sqrt{5}} {\sqrt{5}}$$

The reason we multiplied the denominator by square-root of 5 is because we want to make the denominator a perfect square.

 $$Notice \ also \ that \ \frac{\sqrt{5}} {\sqrt{5}} = 1$$

Therefore, it is like multiplying the expression by 1 which does not change the problem.

 $$\frac{3} {\sqrt{5}} = \frac{3} {\sqrt{5}} × \frac{\sqrt{5}} {\sqrt{5}}$$
 $$\frac{3} {\sqrt{5}} = \frac{3 \sqrt{5}} {\sqrt{25}}$$
 $$\frac{3} {\sqrt{5}} = \frac{3 \sqrt{5}} {5}$$

Example #2

 $$Rationalize \ \frac{ \sqrt{2}} {\sqrt{8n}}$$
 $$Multiply \ by \ \frac{\sqrt{2n}} {\sqrt{2n}}$$

Notice that if you had multiplied by square-root(8n), it will still be correct. Multiplying by square-root(2n) will give you smaller number to deal with though and that is better.

 $$\frac{\sqrt{2}} {\sqrt{8n}} = \frac{\sqrt{2}} {\sqrt{8n}} × \frac{\sqrt{2n}} {\sqrt{2n}}$$
 $$\frac{\sqrt{2}} {\sqrt{8n}} = \frac{\sqrt{2 × 2} × \sqrt{n}} {\sqrt{8n × 2n}}$$
 $$\frac{\sqrt{2}} {\sqrt{8n}} = \frac{\sqrt{4} × \sqrt{n}} {\sqrt{16n^2}}$$
 $$\frac{\sqrt{2}} {\sqrt{8n}} = \frac{2 × \sqrt{n}} {4n}$$
 $$\frac{\sqrt{2}} {\sqrt{8n}} = \frac{\sqrt{n}} {2n}$$

Rationalizing a denominator using conjugates

 $$Rationalize \ \frac{6} {\sqrt{5} - \sqrt{2}}$$
 $$Multiply \ by \ \ \frac{\sqrt{5} + \sqrt{2}} {\sqrt{5} + \sqrt{2}}$$
 $$\frac{6} {\sqrt{5} - \sqrt{2}} = \frac{6} {\sqrt{5} - \sqrt{2}} × \frac{\sqrt{5} + \sqrt{2}} {\sqrt{5} + \sqrt{2}}$$
 $$\frac{6} {\sqrt{5} - \sqrt{2}} = \frac{6 × (\sqrt{5} + \sqrt{2}) } { (\sqrt{5} - \sqrt{2})× (\sqrt{5} + \sqrt{2})}$$
 $$\frac{6} {\sqrt{5} - \sqrt{2}} = \frac{6 × (\sqrt{5} + \sqrt{2}) } { (\sqrt{5})^2 - (\sqrt{2})^2}$$
 $$\frac{6} {\sqrt{5} - \sqrt{2}} = \frac{6 × (\sqrt{5} + \sqrt{2}) } { 5 - 2}$$
 $$\frac{6} {\sqrt{5} - \sqrt{2}} = \frac{6 × (\sqrt{5} + \sqrt{2}) } { 2}$$
 $$\frac{6} {\sqrt{5} - \sqrt{2}} = 3 × (\sqrt{5} + \sqrt{2})$$

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