Question

if x= root 3-root2/root3+root2 and y= root 3+ root 2 / root 3- root 2 find the value of x square +y square +xy

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.

Answer 2

Answer:

x² +y² +xy = 49-20√6+49+20√6 +1

=49+49+1=99

x² +y²+xy is 99

Step-by-step explanation:

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

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

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

= √3² -√2² / √3² -√2²

= 3-2 / 3-2

= 1

x² = $\frac{\sqrt{3}-\sqrt{2} }{\sqrt{3}+\sqrt{2} }$ × $\frac{\sqrt{3}-\sqrt{2} }{\sqrt{3}+\sqrt{2} }$

= (√3 -√2 )² /(√3+√2)²

= 3+2-2√3√2 / 3+2 +2√3√2

=5-2√6 / 5+2√6

= 5-2√6 /5+2√6    ×  5-2√6/ 5-2√6

(5-2√6)² / (5+2√6)(5-2√6)

= 25+24-20√ 6 /25-24

49-20√6

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

= (√3+√2)² /(√3-√2)²

= 3+2+2√3√2 /3+2-2√3√2

=5+2√6 / 5- 2√6

=5+2√6/5-2√6   ×  5+2√6 /5+2√6

( 5+2√6)² / (5+2√6) (5 -2√6)

25+24+20√6 / 25-24

= 49+20√ 6

x² +y² +xy = 49-20√6+49+20√6 +1

=49+49+1=99

x² +y²+xy is 99

#SPJ2