Question

if x is equal to under root 3 + under root 2 upon under root 3 minus under root 2 and Y is equal to under root 3 minus under root 2 upon under root 3 + under root 2 then find the value of x square+ Y square + x y

Answer 1

Given data
x= (√3 + √2) / √3 -√2
y= (√3 - √2 ) / √3+ √2

Required
X^2 + y^2 + x y

please find the attached file
Answer is 99..

Hope it helps you.

Answer 2

Concept:

We need to first recall the concept of Rationalisation to solve this question.

The expressions which contain square roots in denominators, have to be rationalise first to make the denominator free from square root and make the calculation easy.

To rationalise it we multiply both numerator and denominator by an irrational number which is called rationalisation factor.

Given:

The expression for x and y are:

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

To find :

The value of the expression: $x^{2}+y^{2}+xy$ .

Solution:

We have to first rationalise x and y to make the calculation easy .

so,      $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \; \times\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \; and \; y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\; \times\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

          $x=\frac{(\sqrt{3}+\sqrt{2})^{2}}{3-2} \;$              $and \; y=\frac{(\sqrt{3}-\sqrt{2})^{2}}{3-2}$    

          $x=(3+2+2\sqrt{6})$       $and\; y=(3+2-2\sqrt{6})$

                     $and\; y=(5-2\sqrt{6})$

Now, $x^{2} =(5+2\sqrt{6})^{2}$          $and \;y^{2} =(5-2\sqrt{6})^{2}$

         $x^{2} =(25+24+10\sqrt{6})$$and \;y^{2} =(25+24-10\sqrt{6})$

         $x^{2} =(49+10\sqrt{6}) \; \;$        $and \;y^{2} =(49-10\sqrt{6})$

and  $xy=(5+2\sqrt{6})(5-2\sqrt{6})$

        $xy=(25-24)$

        $xy = 1$

Hence, $x^{2}+y^{2}+xy = 49+10\sqrt{6}+49-10\sqrt{6}+1$

                                 = 99