Question

Rationalise the denominator of 1/(2√5 + √3)​

Answer 1

Answer:

$\bf ➤ \underline{Solution-}$

$\sf \: \: \: \dfrac{1}{ 2 \sqrt{5} + \sqrt{3} }$

On rationalising,

$\sf \implies\dfrac{1}{ 2 \sqrt{5} + \sqrt{3} } \times \dfrac{2 \sqrt{5} - \sqrt{3}}{2 \sqrt{5} - \sqrt{3}}$

Combine the fractions,

$\sf \implies \dfrac{1(2 \sqrt{5} - \sqrt{3})}{(2 \sqrt{5} + \sqrt{3})(2 \sqrt{5} - \sqrt{3})}$

We know that,

$\sf \implies (a + b)(a - b) = (a)^{2} - (b)^{2}$

So,

$\sf \implies \dfrac{1(2 \sqrt{5} - \sqrt{3})}{(2 \sqrt{5} )^{2} - (\sqrt{3})^{2} }$

$\sf \implies \dfrac{1(2 \sqrt{5} - \sqrt{3})}{20 - 3 }$

$\sf \implies \dfrac{1(2 \sqrt{5} - \sqrt{3})}{17 }$

$\sf \implies \dfrac{2 \sqrt{5} - \sqrt{3}}{17 }$

Hence,

On rationalising we got,

$\bf \implies\dfrac{2 \sqrt{5} - \sqrt{3}}{17 }$