Question

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

Answer 1

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Q:-solve this and rationalise the denominator

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

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

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

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

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

$⟹ \frac{49 + 6 + 14 \sqrt{6} }{43}$

$⟹ \frac{55 + 14 \sqrt{6} }{43} ✓$

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Answer 2

Answer:-

Rationalising the denominator:

$\bf \implies \: \frac{7 + \sqrt{6} }{7 - \sqrt{6} } \\ \\ \bf \implies \: \frac{7 + \sqrt{6} }{7 - \sqrt{6} } \times \frac{7 + \sqrt{6} }{7 + \sqrt{6} } \\ \\ \bf \implies \: \frac{ {(7 + \sqrt{6} })^{2} }{ {(7)}^{2} - {( \sqrt{6} })^{2} } \\ \\ \bf \implies \: \frac{ {(7)}^{2} + 2(7)( \sqrt{6} ) + {( \sqrt{6} })^{2} }{</u><u>4</u><u>9</u><u> - 6} \\ \\ \bf \implies \: \frac{49 + 14 \sqrt{6} + 6}{</u><u>4</u><u>3</u><u>} \\ \\ \bf \implies \:55 + 14 \sqrt{6} </u><u>{</u><u>4</u><u>3</u><u>}</u><u>$