Question

Rationalisie the denominator of: √18/3-√12​

Answer 1

Answer:

$\longmapsto \: 3 \sqrt{2} - 2 \sqrt{6} .$

Step-by-step explanation:

$\sf{\dfrac{\sqrt{18}}{3 - \sqrt{12}}}$

By Rationalizing the denominator:-

$= \sf{\dfrac{\sqrt{18}}{3 - \sqrt{12}} \times \dfrac{3 + \sqrt{12}}{3 + \sqrt{12}}}$

$= \sf{\dfrac{\sqrt{18}(3 + \sqrt{12})}{(3)^2 - (\sqrt{12})^2}}$

$= \sf{\dfrac{\sqrt{18}(3 + \sqrt{12})}{9 - 12}}$

$= \sf{\dfrac{\sqrt{18}(3 + \sqrt{12})}{-3}}$

$= \sf{\dfrac{3\sqrt{2}(3 + \sqrt{12})}{-3}}$

$= \sf{\dfrac{\not{3\sqrt{2}}(3 + \sqrt{12})}{\not{-3}}}$

$= \sf{-\sqrt{2}(3 + 2\sqrt{3})}$

$= \sf{-3\sqrt{2} - 2\sqrt{6}}$

$\therefore \sf{\dfrac{\sqrt{18}}{3 - \sqrt{12}} = -3\sqrt{2} - 2\sqrt{6}}$