Question

4. Rationalise the denominator 1/3+under root 2

Answer 1

$\sf{{\dfrac{1}{3 + {\sqrt{2}}}}}$


$\sf{\implies {\dfrac{1}{3 + {\sqrt{2}}}} × {\dfrac{3 - {\sqrt{2}}}{3 - {\sqrt{2}}}}}$


$\sf{\implies{\dfrac{(1)(3 - {\sqrt{2}})}{(3 + {\sqrt{2}})(3 - {\sqrt{2}})}}}$


${\boxed{\sf{(a + b )(a - b) = a^2 - b^2}}}$


$\sf{\implies {\dfrac{3 - {\sqrt{2}}}{(3)^2 - ({\sqrt{2}})^2}}}$


$\sf{\implies {\dfrac{3 - {\sqrt{2}}}{9 - 2}}}$


$\sf\red{\implies {\dfrac{3 - {\sqrt{2}}}{7}}}$

Answer 2

$\textbf{\textit{\underline{\huge{Rationalize :}}}}$

$\bold{= \frac{1}{3 + \sqrt{2} }}$

$\textsf{\underline{To Rationalize change the sign}}$
$\textsf{\underline{in denominator}}$

$\bold{= \frac{1}{3 + \sqrt{2} } \times \frac{3 - \sqrt{2} }{3 - \sqrt{2} }} \\ \\ \bold{= \frac{1(3 - \sqrt{2}) }{(3 + \sqrt{2} )(3 - \sqrt{2}) }} \\ \\ \bold{ (a+b)(a-b)=a^2 - b^2 } \\ \\ \bold{= \frac{3 - \sqrt{2} }{ {3}^{2} - { \sqrt{2} }^{2} }} \\ \\ \bold{= \frac{3 - \sqrt{2} }{9 - 2}} \\ \\ \bold{= \frac{3 - \sqrt{2} }{7}}$

$\Longrightarrow{\boxed{\bold{\huge{ \frac{3 - \sqrt{2} }{7}}}}}$