Question

(a) Rationalise the denominator of:
√2/4√2-2√3​

Answer 1

Given :

$\boxed{\bf\dfrac{\sqrt{2}}{4\sqrt{2}-2\sqrt{3}}}$

To Find :

Rationalise the denominator.

Solution :

$\\ =\sf\dfrac{\sqrt{2}}{4\sqrt{2}-2\sqrt{3}}$

$\\ =\sf\dfrac{\sqrt{2}}{4\sqrt{2}-2\sqrt{3}}\times\dfrac{4\sqrt{2}+2\sqrt{3}}{4\sqrt{2}+2\sqrt{3}}$

Using the identity (a + b)(a - b) = a² - b²

$\\ =\sf\dfrac{\sqrt{2}(4\sqrt{2}+2\sqrt{3})}{(4\sqrt{2})^2-(2\sqrt{3})^2}$

$\\ =\sf\dfrac{4(\sqrt{2})^2+2\sqrt{3}\times\sqrt{2}}{(16\times2)-(4\times3)}$

$\\ =\sf\dfrac{(4\times2)+2\sqrt{6}}{32-12}$

$\\ =\sf\dfrac{(4\times2)+2\sqrt{6}}{20}$

$\\ =\sf\dfrac{8+2\sqrt{6}}{20}$

Taking 2 as common in numerator and denominator,

$\\ =\sf\dfrac{2(4+\sqrt{6})}{20}$

$\\ =\sf\dfrac{\not{2}(4+\sqrt{6})}{\cancel{20}\ \ ^{10}}$

$\\ =\sf\dfrac{4+\sqrt{6}}{10}$

$\\ \boxed{\bf\dfrac{4+\sqrt{6}}{10}.}$

The answer is (4 + √6)/10.

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