Question

if x=root 3+root2/root3-root2 then find the value of x2

Answer 1

Answer:

Value of x² + y² + xy is 99.

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 + xy

First we find,

xy=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}

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

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

=\frac{3-2}{3-2}

= 1

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 + xy = 49 - 20√6 + 49 + 20√6 + 1 = 49 + 49 + 1 = 99

Therefore, Value of x² + y² + xy is 99.