Question

Rationalize the denominator of 5/(√3 - √5)

Answer 1

$\mathbb{ SOLUTION }$ :

For rationalising the denominator, we will multiply the numerator and denominator by conjugate of denominator to remove the radical sign from the denominator.

Denominator of $\frac{5}{ \sqrt{3} - \sqrt{5} }$ is $(\sqrt{3} - \sqrt{5})$

Here, the conjugate of denominator $(\sqrt{3} - \sqrt{5})$ is $(\sqrt{3} + \sqrt{5})$

$= \frac{5}{ \sqrt{3} - \sqrt{5} } \times \frac{ \sqrt{3} + \sqrt{5} }{ \sqrt{3} + \sqrt{5} } \: \: \: (by \: \: rationalisation) \\ \\ \\ = \frac{5( \sqrt{3} + \sqrt{5} ) }{ {( \sqrt{3} )}^{2} - {( \sqrt{5}) }^{2} } \: \: \: \: \: ( \therefore \: (a - b)(a + b) = {a}^{2} - {b}^{2} \: ) \\ \\ \\ = \frac{5( \sqrt{3} + \sqrt{5} ) }{3 - 5} \\ \\ \\ = - \frac{5}{2} ( \sqrt{3} + \sqrt{5} )$

Answer 2

Hey there !!


▶ Q. Rationalize the denominator of $\frac{5}{ \sqrt{3} - \sqrt{5} } .$


▶ Solution :-


$\bf = \frac{5}{ \sqrt{3} - \sqrt{5} } .$

$\bf = \frac{5}{ \sqrt{3} - \sqrt{5} } \times \frac{ \sqrt{3} + \sqrt{5} }{ \sqrt{3} + \sqrt{5} } .$

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



$\bf = \frac{5( \sqrt{3} + \sqrt{5} )}{( { \sqrt{3} )}^{2} - ( { \sqrt{5} )}^{2} } .$


$\bf = \frac{5 \sqrt{3} + 5 \sqrt{5} }{3 - 5} .$


$\bf = \frac{5 \sqrt{3} + 5 \sqrt{5} }{ - 2} .$


$\boxed{\bf = \frac{ - 5 \sqrt{3} - 5 \sqrt{5} }{2} .}$

OR

$\boxed { \bf = \frac{ - 5}{2} ( \sqrt{3} + \sqrt{5} ).}$


✔✔ Hence, it is solved ✅✅.

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THANKS


#BeBrainly.