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

Answer:99

Step-by-step explanation:

as given

$x= \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\\\x^{2}=(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}})^{2}\\\\=\frac{3+2+2\sqrt{6}}{3+2-2\sqrt{6}}\\\\=\frac{5+2\sqrt{6}}{5-2\sqrt{6}}$

and

$y= \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\\\y^{2}=(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}})^{2}\\\\=\frac{3+2-2\sqrt{6}}{3+2+2\sqrt{6}}\\\\=\frac{5-2\sqrt{6}}{5+2\sqrt{6}}$

and xy =1

so the value of

$x^{2}+y^{2}+xy\\\\=\frac{5+2\sqrt{6}}{5-2\sqrt{6}}+\frac{5-2\sqrt{6}}{5+2\sqrt{6}}+1\\\\=\frac{(5+2\sqrt{6})^{2}+(5-2\sqrt{6})^{2}}{(5-2\sqrt{6})*(5+2\sqrt{6})}+1\\\\=\frac{2(25+24)}{25-24}+1\\\\=98+1\\\\=99$

Answer 2

Answer:

$x^2+y^2+xy=99$

Step-by-step explanation:

We are given that,

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

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

Now, it is required to find the value of $x^2+y^2+xy$

So, we have,

$x^2=(\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}})^2\\\\\x^2=\dfrac{3+2+2\sqrt{6}}{3+2-2\sqrt{6}}\\\\\x^2=\dfrac{5+2\sqrt{6}}{5-2\sqrt{6}}\\\\\x^2=\dfrac{5+2\sqrt{6}}{5-2\sqrt{6}}\times \dfrac{5+2\sqrt{6}}{5+2\sqrt{6}}\\\\\x^2=\dfrac{(5+2\sqrt{6})^2}{25-4\times 6}\\\\\x^2=\dfrac{25+4\times 6+20\sqrt{6}}{25-24}\\\\x^2=25+24+20\sqrt{6}\\\\x^2=49+20\sqrt{6}$

Also, we have,

$y^2=(\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}})^2\\\\\y^2=\dfrac{3+2-2\sqrt{6}}{3+2+2\sqrt{6}}\\\\\y^2=\dfrac{5-2\sqrt{6}}{5+2\sqrt{6}}\\\\\y^2=\dfrac{5-2\sqrt{6}}{5+2\sqrt{6}}\times \dfrac{5-2\sqrt{6}}{5-2\sqrt{6}}\\\\\y^2=\dfrac{(5-2\sqrt{6})^2}{25-4\times 6}\\\\\y^2=\dfrac{25+4\times 6-20\sqrt{6}}{25-24}\\\\y^2=25+24-20\sqrt{6}\\\\y^2=49-20\sqrt{6}$

Further, we have,

$xy=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

i.e. $xy=1$

Finally, we get,

$x^2+y^2+xy=49+20\sqrt{6}+49-20\sqrt{6}+1$

i.e.  $x^2+y^2+xy=49+20\sqrt{6}+49-20\sqrt{6}+1$

i.e. $x^2+y^2+xy=49+49+1$

i.e. $x^2+y^2+xy=99$

Thus, the required value is 99.