Question

2/√5+√3+2 , rationalize the denominator.​

Answer 1

$\huge{\bf{\green{\mathfrak{\dag{\underline{\underline{Answer:-}}}}}}}$

Given

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

To find

Rationalization of the given number

Solution

Conjugate of (√5+√3)+2 = (√5+√3)-2

For rationalize the denominator we want to multiple it's conjugate to the numerator and denominator

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

$= \frac{2[( \sqrt{5} + \sqrt{3}) - 2]}{[ {( \sqrt{5} + \sqrt{3})}^{2} - {2}^{2} ]} = \frac{{2[( \sqrt{5} + \sqrt{3}) - 2]}}{ {( \sqrt{5}} )^2 + 2 \times \sqrt{5} \times \sqrt{3} + { (\sqrt{3}) }^{2} }$

$= \frac{2[( \sqrt{5} + \sqrt{3}) - 2]}{5 + 2 \sqrt{8} + 3 - 4} =\frac{2[( \sqrt{5} + \sqrt{3}) - 2]}{4 + 4 \sqrt{2} }$

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

The sum is not yet completed.

Now again we want to rationalize the denominator.

Conjugate of 2+2√2 is 2-2√2

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

$=\frac{2√5+2√3-2√10-2√6+4√2}{2²-(2√2)²}$

$= \frac{2√5+2√3-4-2√10-2√6+4√2}{4-8}\\ = \frac{2(√5+√3-2-√10-√6+2√2)}{-4}$

$= \frac{√5+√3-2-√10-√6+2√2}{-2}$