Question

Rationalize the denominator 3/root3+root5-root2

Answer 1

match the answer...hope it shall help u

Answer 2

Answer:

Answer is $\frac{-1}{4}(3\sqrt{2}-2\sqrt{3}-\sqrt{30})$

Step-by-step explanation:

Given Expression = $\frac{3}{\sqrt{3}+\sqrt{5}-\sqrt{2}}$

We have to rationalize the denominator of given expression.

Consider,

$\frac{3}{\sqrt{3}+\sqrt{5}-\sqrt{2}}$

$\frac{3}{(\sqrt{3}+\sqrt{5})-\sqrt{2}}$

$\frac{3}{(\sqrt{3}+\sqrt{5})-\sqrt{2}}\times\frac{(\sqrt{3}+\sqrt{5})+\sqrt{2}}{(\sqrt{3}+\sqrt{5})+\sqrt{2}}$

$\frac{3((\sqrt{3}+\sqrt{5})+\sqrt{2})}{(\sqrt{3}+\sqrt{5})^2-2}$ (using identity in denominator (a+b)(a-b) = a²-b² )

$\frac{3(\sqrt{3}+\sqrt{5})+\sqrt{2})}{3+5+2\sqrt{3}\sqrt{5}-2}$

$\frac{3(\sqrt{3}+\sqrt{5})+\sqrt{2})}{6+2\sqrt{3}\sqrt{5}}$

$\frac{3}{2}\times\frac{(\sqrt{3}+\sqrt{5})+\sqrt{2})}{3+\sqrt{3}\sqrt{5}}$

$\frac{3}{2}\times\frac{(\sqrt{3}+\sqrt{5}+\sqrt{2})}{3+\sqrt{15}}\times\frac{3-\sqrt{15}}{3-\sqrt{15}}$

$\frac{3}{2}\times\frac{(\sqrt{3}+\sqrt{5}+\sqrt{2})(3-\sqrt{15})}{9-15}$ (using same identity in denominator)

$\frac{3}{2}\times\frac{(\sqrt{3}+\sqrt{5}+\sqrt{2})(3-\sqrt{15})}{-6}$

$\frac{-1}{4}(\sqrt{3}+\sqrt{5}+\sqrt{2})(3-\sqrt{15})$

$\frac{-1}{4}(3\sqrt{2}-2\sqrt{3}-\sqrt{30})$ (after simplifying product)

Therfore, Answer is $\frac{-1}{4}(3\sqrt{2}-2\sqrt{3}-\sqrt{30})$