Question

root 3 + root 2 upon root 3 minus root 2

Answer 1

√3+√2/√3-√2


multiply and divide with √3+√2


=(√3+√2)²/(√3)²-(√2)²

=(√3+√2)²/3-2

=(√3+√2)²/1
=3+2+2√6

=5+2√6



I hope this will help u ;)

Answer 2

Given,

root 3 + root 2 upon root 3 minus root 2

To Find,

The result =?

Solution,

We can solve the question as follows:

It is given that we have to find the result of  $\frac{\sqrt{3} +\sqrt{2} }{\sqrt{3} -\sqrt{2} }$.

To solve the given equation, we will first multiply and divide the numerator and denominator with the conjugate of the denominator.

Therefore,

$= \frac{\sqrt{3} + \sqrt{2} }{\sqrt{3}- \sqrt{2} } *\frac{\sqrt{3} + \sqrt{2} }{\sqrt{3} +\sqrt{2} }$

Since, $a^{2} - b^{2} = (a + b)(a - b)$ and $(a + b)*(a + b) = (a +b)^{2}$,

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

$= \frac{(\sqrt{3})^{2} + (\sqrt{2})^{2} +2*\sqrt{3}* \sqrt{2}}{3 - 2 }$

$= \frac{3 + 2 +2*\sqrt{6}}{ 1}$

$= \frac{5 + 2\sqrt{6}}{1}$

Hence, the result is 5 + 2√6.