Question

x=squareroot 3+squareroot 2/squareroot 3 -squareroot 2 and y=squareroot3-squareroot2/squareroot3+ squareroot2 find value of x^2-y^2+xylophone if squareroot6 is 2.4

Answer 1

${(x + y)}^{2} - {(x - y)}^{2} = 4xy \\ {(x + y)}^{2} ={ x}^{2} + {y}^{2} + 2xy\\ x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ x = \frac{{ (\sqrt{3} + \sqrt{2})}^{2} }{ 3 - 2 } = {( \sqrt{3} + \sqrt{2} )}^{2} \\ y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ y = {( \sqrt{3} - \sqrt{2}) }^{2} \\ now \\ {x}^{2} = {( \sqrt{3} + \sqrt{2}) }^{4} \\ {y}^{2} = {( \sqrt{3} - \sqrt{2}) }^{4} \\ xy = {( \sqrt{3} + \sqrt{2} ) }^{2} {( \sqrt{3} - \sqrt{2}) }^{2} = {1}^{2} = 1 \\ now \\ {x}^{2} - {y}^{2} + xy \\ = {( \sqrt{3} + \sqrt{2}) }^{4} - {( \sqrt{3} - \sqrt{2}) }^{4} + 1 \\ = {(5 + 2 \sqrt{6} )}^{2} - {(5 - 2 \sqrt{6} )} ^{2} \\ = 8 \sqrt{6} \\ = 8(2.4) \\ = 19.2$
hope it helps.