MCQ
If $x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ and $y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$, then $x^2+x y+y^2=$
- A$102$
- B$101$
- ✓$99$
- D$98$
✓
Answer
Correct option: C.
$99$
Given $x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ and $y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$,
Consider,
$x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$=\frac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3})^2+(\sqrt{2})^2}$
$=\frac{(\sqrt{3})^2+(\sqrt{2})^2-2(\sqrt{3})(\sqrt{2})}{3-2}$
$=\frac{3+2-2 \sqrt{6}}{1}$
$=5-2 \sqrt{6}$
Hence $x=5-2 \sqrt{6}$
$\Rightarrow x^2=(5-2 \sqrt{6})^2$
$=25+24-20 \sqrt{6}$
$=49-20 \sqrt{6}$
i.e. $x^2=49-20 \sqrt{6}........(i)$
Again consider
$ y =\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$=\frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}$
$=\frac{(\sqrt{3})^2+(\sqrt{2})^2+2(\sqrt{3})(\sqrt{2})}{3-2}$
$=\frac{3+2+2 \sqrt{6}}{1}$
$=5+2 \sqrt{6}$
Hence $y=5+2 \sqrt{6}$
$\Rightarrow y^2=(5+2 \sqrt{6})^2$
$=25+24+20 \sqrt{6}$
$=49+20 \sqrt{6}$
i.e. $y ^2=49+20 \sqrt{6}-\ldots (ii)$
Then $x^2+x y+y^2$
$=49-20 \sqrt{6}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+49+20 \sqrt{6}[$ from $(i)$ nd $(ii)]$
$=98+1$
$=99$
Consider,
$x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$=\frac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3})^2+(\sqrt{2})^2}$
$=\frac{(\sqrt{3})^2+(\sqrt{2})^2-2(\sqrt{3})(\sqrt{2})}{3-2}$
$=\frac{3+2-2 \sqrt{6}}{1}$
$=5-2 \sqrt{6}$
Hence $x=5-2 \sqrt{6}$
$\Rightarrow x^2=(5-2 \sqrt{6})^2$
$=25+24-20 \sqrt{6}$
$=49-20 \sqrt{6}$
i.e. $x^2=49-20 \sqrt{6}........(i)$
Again consider
$ y =\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$=\frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}$
$=\frac{(\sqrt{3})^2+(\sqrt{2})^2+2(\sqrt{3})(\sqrt{2})}{3-2}$
$=\frac{3+2+2 \sqrt{6}}{1}$
$=5+2 \sqrt{6}$
Hence $y=5+2 \sqrt{6}$
$\Rightarrow y^2=(5+2 \sqrt{6})^2$
$=25+24+20 \sqrt{6}$
$=49+20 \sqrt{6}$
i.e. $y ^2=49+20 \sqrt{6}-\ldots (ii)$
Then $x^2+x y+y^2$
$=49-20 \sqrt{6}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+49+20 \sqrt{6}[$ from $(i)$ nd $(ii)]$
$=98+1$
$=99$
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