Question

If x =root3-root2 then find the value of x-1/xwhole square

Answer 1

$\huge\mathbb\red{SOLUTION }$

$\bf\ \: x = \sqrt{3} - \sqrt{2} \\ \\ \bf\ \implies \frac{1}{x} = \frac{1}{ \sqrt{3} - \sqrt{2} } \\ \\ \huge\mathbb{ \pink{TO \: FIND \leadsto }} \\ \\ \bf\boxed{ \green {(x - \frac{1}{x} )}^{2} =? }\\ \\ \bf\ \implies{{( \sqrt{3} - \sqrt{2} - \frac{1}{ \sqrt{3} - \sqrt{2} } })^{2}} \\ \\ \bf\ \implies \frac{ {( \sqrt{3} - \sqrt{2} )}^{2} - 1 }{ \sqrt{3} - \sqrt{2} } \\ \\ \bf\ \implies \frac{3 - 2 \sqrt{6} + 2 - 1}{ \sqrt{3} - \sqrt{2} } \\ \\ \bf\ \implies \frac{ - 2 \sqrt{6} }{ \sqrt{3} - \sqrt{2} } \\ \\ \bf\ \implies \frac{ - 2 \sqrt{6} ( \sqrt{3} + \sqrt{2} )}{( \sqrt{3} - \sqrt{2} )( \sqrt{3} + \sqrt{2} ) } \\ \\ \bf\ \implies \frac{ - 6 \sqrt{2} - 4 \sqrt{3} }{ ({ \sqrt{3} )}^{2} - {( \sqrt{2}) }^{2} } \\ \\ \bf\boxed{ \implies \frac{ - 2(3 \sqrt{2} + 2 \sqrt{3}) }{1} }$