Simplify:$ \frac{7+3 \sqrt{5}}{3+\sqrt{5}}-\frac{7-3 \sqrt{5}}{3-\sqrt{5}} $

Given:

\( \frac{7+3 \sqrt{5}}{3+\sqrt{5}}-\frac{7-3 \sqrt{5}}{3-\sqrt{5}} \)

To do: 

We have to simplify the given expression.

Solution:

We know that,

Rationalising factor of a fraction with denominator ${\sqrt{a}}$ is ${\sqrt{a}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}-\sqrt{b}}$ is ${\sqrt{a}+\sqrt{b}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}+\sqrt{b}}$ is ${\sqrt{a}-\sqrt{b}}$.

Therefore,

$\frac{7+3 \sqrt{5}}{3+\sqrt{5}}-\frac{7-3 \sqrt{5}}{3-\sqrt{5}}=\frac{(7+3 \sqrt{5})(3-\sqrt{5})-(7-3 \sqrt{5})(3+\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}$

$=\frac{(21-7 \sqrt{5}+9 \sqrt{5}-3 \sqrt{5} \times \sqrt{5})-(21+7 \sqrt{5}-9 \sqrt{5}-3 \sqrt{5} \times \sqrt{5})}{(3)^{2}-(\sqrt{5})^{2}}$

$=\frac{(21+2 \sqrt{5}-15)-(21-2 \sqrt{5}-15)}{9-5}$

$=\frac{(6+2 \sqrt{5})-(6-2 \sqrt{5})}{4}$

$=\frac{6+2 \sqrt{5}-6+2 \sqrt{5}}{4}$

$=\frac{4 \sqrt{5}}{4}$

$=\sqrt{5}$

Hence, $\frac{7+3 \sqrt{5}}{3+\sqrt{5}}-\frac{7-3 \sqrt{5}}{3-\sqrt{5}}=\sqrt5$. 

Tutorialspoint

Updated on: 10-Oct-2022

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