Question

Rationalise the denominator.

$\frac{ \sqrt{9} }{ \sqrt{8} - \sqrt{5} }$

Answer 1

Hola Mate!!

Your answer :-

$= > \frac{ \sqrt{9} }{ \sqrt{8} - \sqrt{5} } \\ \\ = > \frac{ \sqrt{3 \times 3} }{ \sqrt{2 \times 2 \times 2} - \sqrt{5} } \\ \\ = > \frac{3}{2 \sqrt{2} - \sqrt{5} } \\ \\ = > \frac{3}{2 \sqrt{2} - \sqrt{5} } \times \frac{2 \sqrt{2} + \sqrt{5} }{2 \sqrt{2} + \sqrt{5} } \\ \\ = > \frac{3(2 \sqrt{2} + \sqrt{5} )}{ {(2 \sqrt{2}) }^{2} - {( \sqrt{5}) }^{2} } \\ \\ = > \frac{6 \sqrt{2} + 3 \sqrt{5} }{8 - 5} \\ \\ = > \frac{6 \sqrt{2} + 3 \sqrt{5} }{3} \\ \\ = > \frac{6 \sqrt{2} }{3} + \frac{3 \sqrt{5} }{3} \\ \\ = > 2 \sqrt{2} + \sqrt{5}$

☆ Hope it helps ☆

Answer 2

$\dfrac{\sqrt{9}}{\sqrt{8} - \sqrt{5}}$

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Rationalise the denominator:

$= \dfrac{\sqrt{9}}{\sqrt{8} - \sqrt{5}} \times \dfrac{\sqrt{8} + \sqrt{5}}{\sqrt{8} + \sqrt{5}}$

$= \dfrac{\sqrt{9} (\sqrt{8} + \sqrt{5} )}{8 - 5}$

$= \dfrac{3 (2\sqrt{2} + \sqrt{5} )}{3}$

$= 2\sqrt{2} + \sqrt{5}$

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Answer: 2√2 + √5