Question

ifx=root3--root2/root3+root2,y=root3+root2/root3-root2,find the value of xsquare +xy+ysquare

Answer 1

Hello,

$x=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} =\dfrac{(\sqrt{3}-\sqrt{2})^2}{3-2} =3+2-2\sqrt{6} =5-2\sqrt{6} \\ y=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} =\dfrac{(\sqrt{3}+\sqrt{2})^2}{3-2} =3+2+2\sqrt{6} =5+2\sqrt{6} \\ x^2+y^2+xy=(5-2\sqrt{6})^2+(5+2\sqrt{6})^2+(5-2\sqrt{6})*(5+2\sqrt{6})\\ =25+4*6-20\sqrt{6}+25+4*6+20\sqrt{6}+25-4*6\\ =2*49+1=99$

Answer 2

Answer:

$\text{The value of } x^2+ y^2 + xy\text{ is 99.}$

Step-by-step explanation:

Given that

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

$\text{we have to find the value of } x^2+ y^2 + xy$

First we find the value of xy

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

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

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

$=\frac{3-2}{3-2}=1$

$\text{now, we will find the value of }x^2$

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

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

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

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

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

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

$=\frac{25+24-20\sqrt{6}}{25-24}$  

$49-20\sqrt6$

$\text{now, we will find the value of }y^2$

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

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

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

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

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

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

$=\frac{25+24+20\sqrt{6}}{25-24}$  

$49+20\sqrt6$

Now add all three terms, we get

$x^2+y^2+xy = 49 - 20\sqrt6 + 49 + 20\sqrt6 + 1 = 49 + 49 + 1 = 99$

$\text{Hence, the value of } x^2+ y^2 + xy\text{ is 99.}$