Question

Rationalisation of denominator
$\frac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} }$
​

Answer 1

$\frac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} } \\ = \frac{ \sqrt{6}( \sqrt{2} - \sqrt{3}) }{( \sqrt{2} + \sqrt{3})( \sqrt{2} - \sqrt{3} ) } \\ = \frac{ \sqrt{12} - \sqrt{18} }{( \sqrt{2}) {}^{2} - ( \sqrt{3}) {}^{2} } \\ = \frac{2 \sqrt{3} - 3 \sqrt{2} }{2 - 3} \\ = \frac{ - \sqrt{2} }{ - 1} \\ = \sqrt{2}$

Answer 2

Given :

$\boxed{\bf \dfrac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} } }$

To Find :

Rationalization of the denominator.

Solution :

$\\ \sf=\dfrac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

Rationalizing the denominator,

$\\ \sf=\dfrac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\times\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}$

$\\ \sf=\dfrac{(\sqrt{6})\times(\sqrt{2}-\sqrt{3})}{(\sqrt{2}+\sqrt{3})\times(\sqrt{2}-\sqrt{3})}$

$\\ \sf=\dfrac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{(\sqrt{2}+\sqrt{3})\times(\sqrt{2}-\sqrt{3})}$

Using identity (a + b)(a - b) =(a² - b²) in denominator,

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

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

$\\ \sf=\dfrac{\sqrt{12}-\sqrt{18}}{-1}$

$\\ \sf=\dfrac{-(\sqrt{12}-\sqrt{18})}{1}$

$\\ \sf=\dfrac{-\sqrt{12}+\sqrt{18}}{1}$

$\\ \sf=\sqrt{18}-\sqrt{12}$

$\\ \therefore\boxed{\bf\sqrt{18}-\sqrt{12}.}$

The answer is √18 - √12.

Explore More :

• The above question is solved by the method of rationalisation.

• Rationalization is a technique which is generally used to remove the roots which is given in the form of denominators .

• We use rationalisation for proving something like in trigonometry.

• For doing rationalising you always have to multiply the same value but having opposite sign.