Question

1/
$\sqrt{3 - \sqrt{2} }$
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Answer 1

$\huge \purple {\sf {\frac{1}{ \sqrt{3} - \sqrt{2} }} }$

$\huge \purple {\sf {= \frac{1}{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} }}}$

$\huge \purple {\sf {= \frac{ \sqrt{3} + \sqrt{2} }{3 - 2}}}$

$\huge \purple {\sf {= \frac{ \sqrt{3} + \sqrt{2} }{1}}}$

$\huge \purple {\sf {= { \sqrt{3} + \sqrt{2} }}}$

Answer 2

Answer:

$\huge\sf{ \red{ \underline{{given \ \: : }}}}$

$\huge\bf \frac{1}{3 - \sqrt{2} }$

$\huge{ \sf{ \red{ \underline{to \: find}}}}$

$\bf the \: rationaling\:factor$

$\huge \sf \red{ \underline{{solutio n\ \: : }}}$

$\huge \bf{ \boxed{ \frac{1}{3 - \sqrt{2} } }}$

$\huge \to \sf \frac{1 \times }{3 - \sqrt{2} \times } \frac{3 + \sqrt{2} }{3 + \sqrt{2} }$

$\huge\to \sf \frac{3 + \sqrt{2} }{ {3}^{2} - { \sqrt{2} }^{2} }$

$\huge\sf \to \frac{3 + \sqrt{2} }{9 - 2}$

${ \boxed{\red \to{ \sf \frac{3 + \sqrt{2} }{7} }}}$

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Identity used:-

$\sf(x + y)(x - y) = {x}^{2} - {y}^{2}$

Explanation:-

• To cancel a root term we need to square it.

• By using this identity we can square both the values.

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