Question

Rationalize the denominator of 1/3- square root5​

Answer 1

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

It is given that we have to rationalize the given number's denominator that is

${\small{\underline{\boxed{\sf{\dfrac{1}{3-\sqrt{5}}}}}}}$

Now let us rationalize.

${\small{\underline{\boxed{\sf{\dfrac{1}{3-\sqrt{5}}}}}}} \\ \\ :\implies \sf \dfrac{1}{3-\sqrt{5}} \\ \\ :\implies \sf \dfrac{1}{3-\sqrt{5}} \times \dfrac{3+\sqrt{5}}{3+\sqrt{5}} \\ \\ \sf Using \: identity \: is \: mentioned \: below \\ \\ \sf (a-b)(a+b) = a^2 - b^2 \\ \\ \sf According \: to \: the \: identity \\ \\ \sf Here, \: a \: is \: 3 \: and \: b \: is \: \sqrt{5} \\ \\ \sf Putting \: the \: value \: according \: to \: identity \\ \\ :\implies \sf \dfrac{3+\sqrt{5}}{(3)^{2} - (\sqrt{5})^{2}} \\ \\ :\implies \sf \dfrac{3+\sqrt{5}}{9 - 5} \\ \\ :\implies \sf \dfrac{3+\sqrt{5}}{4} \\ \\ \sf Therefore, \: we \: get \: our \: required \: solution. \\ \\ {\small{\underline{\boxed{\sf{\dfrac{3+\sqrt{5}}{4}}}}}} \\ \\ {\pmb{\sf{Henceforth, \: Rationalize!}}}$

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

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

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

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

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

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

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