Question

Rationalise the denominator

$\boxed{\rm \dfrac{6}{\sqrt{8}-3}}$​

Answer 1

$\Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}$

• $\sf \dfrac{6}{\sqrt{8} \: - \: 3}$

$\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Find:-}}}}}}}$

• Rationalise the denominator

$\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Solution:-}}}}}}}$

• Rationalising factor of denominator i.e √8 - 3 is √8 + 3.

Multiplying numerator and denominator with √8 + 3,

• $\sf \dfrac{6}{\sqrt{8} \: - \: 3} \: * \: \dfrac{\sqrt{8} \: + \: 3}{\sqrt{8} \: + \: 3}$

• $\sf \dfrac{(6)(\sqrt{8} \: + \: 3)}{(\sqrt{8} \: - \: 3)(\sqrt{8} \: + \: 3)}$

Here,

In the denominator we are using,

• (a - b)(a + b) = a² - b²

• $\sf \dfrac{6\sqrt{8} \: + \: 18}{(\sqrt{8})^2 \: - \: (3)^2}$

• $\sf \dfrac{6\sqrt{8} \: + \: 18}{8 \: - \: 9}$

• $\sf \dfrac{6\sqrt{8} \: + \: 18}{ - \: 1}$

• $\sf - \: \bigg(\dfrac{6\sqrt{8} \: + \: 18}{1}\bigg)$

• $\sf - \: (6\sqrt{8} \: + \: 18)$

• $\sf - \: 6\sqrt{8} \: - \: 18$

Therefore,

• $\boxed{\pink{\sf \dfrac{6}{\sqrt{8} \: - \: 3} \: = \: - \: 6\sqrt{8} \: - \: 18}}$

Answer 2

${\bigstar \:{\large{\pmb{\sf{\underline{RequirEd \; Solution...}}}}}}$

${\small{\underline{\boxed{\pmb{\sf{Rationalize \: the \: denominator \: = \dfrac{6}{\sqrt{8}-3}}}}}}} \\ \\ :\implies {\pmb{\sf{\dfrac{6}{\sqrt{8}-3}}}} \\ \\ :\implies {\pmb{\sf{\dfrac{6}{\sqrt{8}-3} \: \: \cdot \dfrac{\sqrt{8}+3}{\sqrt{8}+3}}}} \\ \\ \sf We \: have \: to \: use \ the \: given \: identity \\ \\ {\small{\underline{\boxed{\pmb{\sf{\red{(a-b) (a+b) \: = a^2 - b^2}}}}}}} \\ \\ \sf According \: to \: the \: identity \\ \\ \sf Here, \: a \: is \: \sqrt{8} \: and \: b \: is \: 3 \\ \\ \sf Firstly, \: let \: us \: solve \: denominator \\ \\ :\implies {\pmb{\sf{(\sqrt{8})^{2} - (3)^{2}}}} \\ \\ :\implies {\pmb{\sf{8 - 9}}} \\ \\ :\implies {\pmb{\sf{-1}}} \\ \\ \sf Now \: let \: us \: carry \: on \\ \\ :\implies {\pmb{\sf{\dfrac{6 \sqrt{8} + 18}{ - 1}}}} \\ \\ :\implies {\pmb{\sf{-6 \sqrt{8} + 18}}}$

Henceforth, we get our solution!

${\bigstar \:{\large{\pmb{\sf{\underline{AdditioNal \; information...}}}}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto \sqrt{ab} \: = \sqrt{a} \sqrt{b}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto \sqrt{\dfrac{a}{b}} \: = \dfrac{\sqrt{a}}{\sqrt{b}}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (\sqrt{a} + \sqrt{b}) (\sqrt{a} - \sqrt{b}) = a - b}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (a+\sqrt{b}) (a-\sqrt{b}) = a^2 - b}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (\sqrt{a} + \sqrt{b}) (\sqrt{c} - \sqrt{d}) = \sqrt{ac} + \sqrt{as} + \sqrt{bc} + \sqrt{bd}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (\sqrt{a} + \sqrt{b})^{2} \: = a + 2\sqrt{ab}+b}}$