Question

X=root 3 +root 2/root3-root 2 and y=root3-root2/ root 3+root2find the value of x square +y square

Answer 1

$x + y = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } + \frac{ \sqrt{3 } - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ = \frac{ {( \sqrt{3} + \sqrt{2}) }^{2} + {( \sqrt{3} - \sqrt{2} ) }^{2} }{( \sqrt{3} + \sqrt{2} )( \sqrt{3} - \sqrt{2} )} \\ = \frac{6 + 4}{3 - 2} = 10$
$xy = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } = 1$
${x}^{2} + {y}^{2} = {(x + y)}^{2} - 2xy \\ = {10}^{2} - 2 \times 1 = 98$

Answer 2

Answer:

Value of x² + y²  is 98.

Step-by-step explanation:

Given:

$x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\:\:and\:\:y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

To find: value of x² + y²

First we find,

$x^2=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

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

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

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

$=\frac{5-2\sqrt{6}}{5+2\sqrt{6}}\times\frac{5-2\sqrt{6}}{5-2\sqrt{6}}$

$=\frac{(5-2\sqrt{6})^2}{(5+2\sqrt{6})(5-2\sqrt{6})}$

$=\frac{25+24-20\sqrt{6}}{25-24}$  

= 49 - 20√6

$y^2=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

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

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

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

$=\frac{5+2\sqrt{6}}{5-2\sqrt{6}}\times\frac{5+2\sqrt{6}}{5+2\sqrt{6}}$

$=\frac{(5+2\sqrt{6})^2}{(5+2\sqrt{6})(5-2\sqrt{6})}$

$=\frac{25+24+20\sqrt{6}}{25-24}$  

= 49 + 20√6

Now,

x² + y² = 49 - 20√6 + 49 + 20√6 = 49 + 49 = 98

Therefore, Value of x² + y² is 98.