Question

rationalize the denominator of 1/2√5-√3​

Answer 1

$\sf\huge {\underline{\underline{{Solution}}}}$

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

$\sf\:Now, \:Multiply \:both \:numerator$$\sf\: and \:denominator \:with\: opposite$$\sf\: sign \:of \:denominator \:term.$

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

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

$\sf\:Here, \:in \:the\: denominator$ $\sf\:an\: identity\: is \:used:$

$\sf \bold {\boxed{(x + y)(x - y) = {x}^{2} - {y}^{2} }}$

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

$\sf\implies \dfrac{2 \sqrt{5} + \sqrt{3} }{20 - 3}$

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

$\sf \therefore \dfrac{2 \sqrt{5} + \sqrt{3} }{17} \: is \: the \: rationalised \: form$

Learn to know =>

$\sf\:Rationalising\: means \:removing$ $\sf\:the\: root\: values\: from$$\sf\: its\: denominator \:and$ $\sf\:shifts\: towards\: the \:numerator.$