Question

Rationalizing the denominator of $\frac { \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} }$

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Answer 1

$\huge \bf{ \green{ \mid{ \overline{ \underline{ANSWER}}} \mid}}$

Rationalizing the denominator of $\frac { \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} }$

The denominator is

$\sqrt{5} - \sqrt{3}$

We know that, Rationalizing factor ( R. F) will be its conjugate.

So, R. F =

$\sqrt{5} + \sqrt{3}$

Now, We shall proceed like this

$= \frac { \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } \\ \\ = \frac { \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } \times \frac { \sqrt{5} + \sqrt{3} }{ \sqrt{5} + \sqrt{3} } \\ \\ = \frac{ (\sqrt{5} + \sqrt{3}) ^{2} }{ (\sqrt{5} ) ^{2} - (\sqrt{3} ) ^{2} } \\ \\ = \frac{5 + 3 + 2 \sqrt{15} }{5 - 3} \\ \\ = \frac{8 + 2 \sqrt{15} }{2} \\ \\ = \frac{2(4 + \sqrt{15}) }{2} \\ \\ = 4 + \sqrt{15}$

Thus, We have rationalized the denominator of $\frac { \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } \:$

Answer 2

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