Question

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

The value of $x^2+xy+y^2$ is

• A. 99
• B. 100
• C. 1
• D. 0

Answer 1

The correct option is A 99

Given, $x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=3+2-2\sqrt{6}=5-2\sqrt{6}$
$y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=3+2+2\sqrt{6}=5+2\sqrt{6}$
$xy=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=1$
Now, $x^2+xy+y^2=(5-2\sqrt{6}{)}^2+1+(5+2\sqrt{6}{)}^2$
$=25+24-20\sqrt{6}+1+25+24+20\sqrt{6}$
$=99$
Hence, option A is correct.