Question

Rationalize the denominator
$\tt(3 \sqrt{2 } - \sqrt{3} )(4 \sqrt{3} - \sqrt{2} )$

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Answer 1

QUESTION :- Rationlize the denominator $\frac{3 \sqrt{2} - \sqrt{3} }{4 \sqrt{3} - \sqrt{2} }$.

[NOTE :- There is mistake in question you have it will be in the form of $\frac{p}{q}$, you have written it in multiplying form]

SOLUTION :-

$\frac{3 \sqrt{2} - \sqrt{3} }{4 \sqrt{3} - \sqrt{2} }$

$= \frac{3 \sqrt{2} - \sqrt{3} }{4 \sqrt{3} - \sqrt{2} } \times \frac{4 \sqrt{3} + \sqrt{2} }{4 \sqrt{3} + \sqrt{2}}$

$= \frac{3 \sqrt{2}(4 \sqrt{3} + \sqrt{2}) - \sqrt{3}(4 \sqrt{3} + \sqrt{2})}{ {(4 \sqrt{3})}^{2} - {(\sqrt{2})}^{2} }$

$= \frac{12 \sqrt{6} + 3 \sqrt{4} - 4 \sqrt{9} - \sqrt{6} }{4 \times 3 - 2}$

$= \frac{12 \sqrt{6} + 3 \times 2 - 4 \times 3 - \sqrt{6} }{12 - 2}$

$= \frac{12 \sqrt{6} + 6 - 12 - \sqrt{6}}{10}$

$= \frac{12 \sqrt{6} - \sqrt{6} - 6}{10}$

$\fbox \orange{= \frac{11 \sqrt{6} - 6}{10}}$

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Hope it helps ✌

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