Question

If $x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}andy=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}},$ then $x^2+xy+y^2=$

• A. 101
• B. 99
• C. 98
• D. 102

Answer 1

The correct option is B

99

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

(Rationalising the denominator)

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

$=\frac{5-2\sqrt{6}}{1}=5-2\sqrt{6}$

Similarly,

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

(Rationalising the denominator)

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

$=\frac{5+2\sqrt{6}}{1}=5+2\sqrt{6}$

$\therefore x^2+y^2+xy=(5-2\sqrt{6}{)}^2+(5+2\sqrt{2}{)}^2+(5-2\sqrt{6})(5+2\sqrt{6})$

$=25+24-20\sqrt{6}+25+24+20\sqrt{6}+(5{)}^2-(2\sqrt{6}{)}^2$

$=49+49+25-24=99$