Question

rationalise the denominator
$\frac{7 + \sqrt{6} }{7 - \sqrt{6} }$
​

Answer 1

$\green{\bold{\underline{ ☆ UPSC-ASPIRANT ☆} }}$

$\red{\bold{\underline{\underline{QUESTION:-}}}}$

Q:-solve this and rationalise the denominator

$\frac{7 + \sqrt{6} }{7 - \sqrt{6} }$

$\huge\tt\underline\blue{ANSWER }$

------>>>>Here is your answer<<<<--------

$⟹ \frac{7 + \sqrt{6} }{7 - \sqrt{6} } \times \frac{7 + \sqrt{6} }{7 + \sqrt{6} }$

$⟹ \frac{ {(7 + \sqrt{6}) }^{2} }{ {(7)}^{2} - {( \sqrt{6}) }^{2} }$

$⟹ \frac{ {(7)}^{2} + {( \sqrt{6} ) }^{2} + 2(7)( \sqrt{6} ) }{49 - 6}$

$⟹ \frac{49 + 6 + 14 \sqrt{6} }{43}$

$⟹ \frac{55 + 14 \sqrt{6} }{43} ✓$

HOPE IT HELPS YOU..

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Thankyou:)

Answer 2

$\bf \huge \underbrace{answer} \\ \\ \tt \: rationalising \: the \: \: denominator \\ \\ \tt \mapsto \: \frac{7 + \sqrt{6} }{7 - \sqrt{6} } \\ \\ \tt \mapsto \: \frac{7 + \sqrt{6} }{7 - \sqrt{6} } \times \frac{7 + \sqrt{6} }{7 + \sqrt{6} } \\ \\ \tt \mapsto \: \frac{ {(7 + \sqrt{6} })^{2} }{ {(7)}^{2} - {( \sqrt{6} })^{2} } \\ \\ \tt \mapsto \: \frac{ {(7)}^{2} + 2(7)( \sqrt{6} ) + {( \sqrt{6} })^{2} }{49 - 6} \\ \\ \tt \mapsto \: \frac{49 + 14 \sqrt{6} + 6}{43} \\ \\ \tt \mapsto \: \frac{55 + 14 \sqrt{6} }{43}$