Question

rationalize the denominator root 2 upon root 2 + root 3 minus root 5​

Answer 1

Answer:

plzzzz give me brainliest ans

Answer 2

Hence, the correct answer is $\frac{\sqrt{6}+3+\sqrt{15} }{6}$.

Given:

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

To Find:

The rationalization of the denominator of $\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$

Solution:

By rationalization, we remove the terms containing roots in the denominator.

The given fraction can be written as

$\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$ = $\frac{\sqrt{2}}{(\sqrt{2}+\sqrt{3})-\sqrt{5}}$ where (√2 + √3) is one term and √5 is the other term of the denominator.

To rationalize the given fraction, we need to multiply the conjugate of the denominator, i.e. (√2 + √3) + √5 to both the numerator and the denominator. Hence, we get:

$\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$ =  $\frac{\sqrt{2}}{(\sqrt{2}+\sqrt{3})-\sqrt{5}}$ x $\frac{(\sqrt{2}+\sqrt{3})+\sqrt{5}}{(\sqrt{2}+\sqrt{3})+\sqrt{5}}$ =  $\frac{2+\sqrt{6}+\sqrt{10} }{(\sqrt{2}+\sqrt{3})^2-\sqrt{5}^2}$

⇒  $\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$ = $\frac{2+\sqrt{6}+\sqrt{10} }{(2+2\sqrt{6}+3)-5}$ = $\frac{2+\sqrt{6}+\sqrt{10} }{(2+2\sqrt{6}+3)-5}$ = $\frac{2+\sqrt{6}+\sqrt{10} }{2\sqrt{6}}$

We observe that we have not reached the rationalized form of the given fraction as there is √6 in the denominator.

Multiplying the numerator and the denominator by √6, we get

 $\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$ = $\frac{2+\sqrt{6}+\sqrt{10} }{2\sqrt{6}}$ x $\frac{\sqrt{6} }{\sqrt{6}}$ = $\frac{2\sqrt{6}+6+2\sqrt{15} }{12}$ = $\frac{\sqrt{6}+3+\sqrt{15} }{6}$.

Hence, the correct answer is $\frac{\sqrt{6}+3+\sqrt{15} }{6}$.

#SPJ2