Question

Rationalize the denominator and simplify √5+√3/3-√5

Answer 1

to rationalize the denominator,we have to multiply denominator and numerator with conjugate of denominator
√5+√3/3-√5=√5+√3×(3+√5)/3-√5×(3+√5)
=3√5+√25+3√3+√15/9-5
=3√5+5+3√3+√15/4
so the denominator is rationalized

Answer 2

FULL ANSWER WITH STEPS

$\frac{\sqrt{5}+\sqrt{3}}{3-\sqrt{5}}$
$= \frac{\sqrt{5}+\sqrt{3}}{3-\sqrt{5}} * \frac{3+\sqrt{5}}{3+\sqrt{5}}$    [Multiplying both numerator and denominator by the conjugate of $3-\sqrt{5}$]
$= \frac{(\sqrt{5}+\sqrt{3})(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}$
$= \frac{\sqrt{5}*3+\sqrt{5}*\sqrt{5}+\sqrt{3}*3+\sqrt{3}*\sqrt{5}}{(3-\sqrt{5})(3+\sqrt{5})}$
$= \frac{3\sqrt{5}+5+3\sqrt{3}+\sqrt{15}}{(3-\sqrt{5})(3+\sqrt{5})}$
$= \frac{3\sqrt{5}+5+3\sqrt{3}+\sqrt{15}}{3^{2}-(\sqrt{5})^{2}}$    [∵ $(a-b)(a+b) = a^{2}-b^{2}$]
$= \frac{3\sqrt{5}+5+3\sqrt{3}+\sqrt{15}}{9-5}$
$= \frac{3\sqrt{5}+5+3\sqrt{3}+\sqrt{15}}{4}$
Congrats ! This is the answer. 

Hope this may help you. 

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