Question

If x=root 3+root2/root3-root2 and y=root3-root2/root3+2 ,find the value of(x+y)square

Answer 1

$\underline{ \large\bf{ question}} : - \\ \sf{ if \: x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \: and \: y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } } \\ \sf{find \: {(x + y)}^{2} } \\ \\ \large \: \underline{\bf{answer}} : - \\ \implies \: \boxed{ \bf{{(x + y)}^{2} = 100}} \: \\ \large \bf{ \underline{explanation \: }} : -$

Now take that which we need ,

$\rightarrow \: \sf{ {(x + y)}^{2} } \\ \\ \rightarrow \sf{ { \bigg( \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } + \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \bigg) }^{2} } \\ \\ \rightarrow \sf { \bigg( \frac{ {( \sqrt{3} + \sqrt{2} )}^{2} + {( \sqrt{3} - \sqrt{2} )}^{2} }{( \sqrt{3} + \sqrt{2})( \sqrt{3} - \sqrt{2} ) } \bigg)}^{2} \\ \\ \because \sf{ (a + b)(a - b) = {a}^{2} - {b}^{2} } \\ \therefore \\ \\ \rightarrow \sf { \bigg( \frac{3 + 2 + 2 \sqrt{6} + 3 + 2 - 2 \sqrt{6} }{3 - 2} \bigg) }^{2} \\ \\ \rightarrow \: { \bigg( \frac{10}{1} \bigg)}^{2} = 100 \\ \\ \sf{ hence} \\ \\ \implies \: \boxed{ \bf{{(x + y)}^{2} = 100}}$

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