MCQ
If $x=\frac{2}{\sqrt{10}-\sqrt{8}}$ and $y=\frac{2}{\sqrt{10}+2 \sqrt{2}}$, then $(x-y)^2=$
- A$4 \sqrt{2}$
- ✓32
- C$8 \sqrt{2}$
- D64
✓
Answer
Correct option: B.
32
(b)
We have,$x=\frac{2}{\sqrt{10}-\sqrt{8}}$ and $y=\frac{2}{\sqrt{10+2 \sqrt{2}}}$
$\begin{array}{ll}\Rightarrow & x=\frac{2}{\sqrt{10}-2 \sqrt{2}} \text { and } y=\frac{2}{\sqrt{10}+2
\sqrt{2}} \\ \Rightarrow & x=\frac{2(\sqrt{10}+2 \sqrt{2})}{(\sqrt{10}+2 \sqrt{2})(\sqrt{10}-2 \sqrt{2})} \text {
and } y=\frac{2(\sqrt{10}-2 \sqrt{2})}{(\sqrt{10}+2 \sqrt{2})(\sqrt{10}-2 \sqrt{2})} \\
\Rightarrow & x=\frac{2(\sqrt{10}+2 \sqrt{2})}{10-8} \text { and } y=\frac{2(\sqrt{10}-2 \sqrt{2})}{10-8} \\
\Rightarrow & x=\sqrt{10}+2 \sqrt{2} \text { and } y=\sqrt{10}-2 \sqrt{2}
\Rightarrow x-y=4 \sqrt{2} \Rightarrow(x-y)^2=32\end{array}$
We have,$x=\frac{2}{\sqrt{10}-\sqrt{8}}$ and $y=\frac{2}{\sqrt{10+2 \sqrt{2}}}$
$\begin{array}{ll}\Rightarrow & x=\frac{2}{\sqrt{10}-2 \sqrt{2}} \text { and } y=\frac{2}{\sqrt{10}+2
\sqrt{2}} \\ \Rightarrow & x=\frac{2(\sqrt{10}+2 \sqrt{2})}{(\sqrt{10}+2 \sqrt{2})(\sqrt{10}-2 \sqrt{2})} \text {
and } y=\frac{2(\sqrt{10}-2 \sqrt{2})}{(\sqrt{10}+2 \sqrt{2})(\sqrt{10}-2 \sqrt{2})} \\
\Rightarrow & x=\frac{2(\sqrt{10}+2 \sqrt{2})}{10-8} \text { and } y=\frac{2(\sqrt{10}-2 \sqrt{2})}{10-8} \\
\Rightarrow & x=\sqrt{10}+2 \sqrt{2} \text { and } y=\sqrt{10}-2 \sqrt{2}
\Rightarrow x-y=4 \sqrt{2} \Rightarrow(x-y)^2=32\end{array}$
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