Question

Rationalise the denominator
(6+4√3)/(6-4√3)

Answer 1

We have to rationalise the denominator of the following expression :-

$\implies \dfrac{6 + 4 \sqrt{3} }{6 - 4 \sqrt{3}}$

To rationalise the denominator multiply numerator and denominator by 6 + 4 √3 :-

$\implies \dfrac{6 + 4 \sqrt{3} }{6 - 4 \sqrt{3}} \times \dfrac{6 + 4 \sqrt{3} }{6 + 4 \sqrt{3}}$

$\implies \dfrac{ (6 + 4 \sqrt{3}) (6 + 4 \sqrt{3})}{(6 - 4 \sqrt{3} )(6 + 4 \sqrt{3})}$

We know that :-

• (a+b) (a+b) = (a+b)²

• (a+b) (a-b) = a²-b²

$\implies \dfrac{ {(6 + 4 \sqrt{3})}^{2} }{( {(6)}^{2} - {(4 \sqrt{3} )}^{2} }$

$\implies \dfrac{ { {(6)}^{2} +(4 \sqrt{3} )}^{2} + 2.6.4 \sqrt{3}}{( {(6)}^{2} - {(4 \sqrt{3} )}^{2} }$

$\implies \dfrac{36 + 48 + 48 \sqrt{3} }{36 - 48}$

$\implies \dfrac{84 + 48 \sqrt{3} }{ - 12}$

Taking out 12 as common from numerator :-

$\implies \dfrac{12(7 + 4 \sqrt{3}) }{ - 12}$

Now cancel out 12 from numerator and denominator :-

$\implies \dfrac{ - 1(7 + 4 \sqrt{3}) }{ 1}$

Therefore we get :-

$\implies - 1(7 + 4 \sqrt{3})$

$\implies - 7 - 4 \sqrt{3}$

Answer :

$- 7 - 4 \sqrt{3}$