MCQ
If $\text{x}=\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$ and $\text{y}=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2},$ then $\text{x}^2+\text{xy}+\text{y}^2=$
- A$101$
- ✓$99$
- C$98$
- D$102$
✓
Answer
Correct option: B.
$99$
$\text{x}=\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$
$\therefore\text{x}=\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\times\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}$
$ =\frac{\big(\sqrt3-\sqrt2\big)^2}{3-2}=5-2\sqrt6$
$\text{y}=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2},$
$\therefore\text{y}=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}$
$ =\frac{\big(\sqrt3+\sqrt{2}\big)^2}{3-2}=5+2\sqrt6$
Now, $\text{x}^2+\text{xy}+\text{y}^2$
$=\big(5 -2\sqrt6\big)^2+\big(5-2\sqrt6\big)\big(5+2\sqrt6\big)+\big(5+2\sqrt6\big)^2$
$=\big(25+24-20\sqrt6\big)+(25-24)+\big(25+24+20\sqrt6\big)$
$=49-20\sqrt6+1+49+20\sqrt6$
$=99$
Hence, correct option is $(b)$.
$\therefore\text{x}=\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\times\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}$
$ =\frac{\big(\sqrt3-\sqrt2\big)^2}{3-2}=5-2\sqrt6$
$\text{y}=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2},$
$\therefore\text{y}=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}$
$ =\frac{\big(\sqrt3+\sqrt{2}\big)^2}{3-2}=5+2\sqrt6$
Now, $\text{x}^2+\text{xy}+\text{y}^2$
$=\big(5 -2\sqrt6\big)^2+\big(5-2\sqrt6\big)\big(5+2\sqrt6\big)+\big(5+2\sqrt6\big)^2$
$=\big(25+24-20\sqrt6\big)+(25-24)+\big(25+24+20\sqrt6\big)$
$=49-20\sqrt6+1+49+20\sqrt6$
$=99$
Hence, correct option is $(b)$.
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