Question : If $x=5–\sqrt{21}$, the value of $\frac{\sqrt{x}}{\sqrt{32–2x}–\sqrt{21}}$ is:
Option 1: $\frac{1}{\sqrt2}(\sqrt{3}–\sqrt{7})$
Option 2: $\frac{1}{\sqrt2}(\sqrt{7}–\sqrt{3})$
Option 3: $\frac{1}{\sqrt2}(\sqrt{7}+\sqrt{3})$
Option 4: $\frac{1}{\sqrt2}(7–\sqrt{3})$
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Correct Answer: $\frac{1}{\sqrt2}(\sqrt{7}–\sqrt{3})$
Solution :
Given: $x=5–\sqrt{21}$
⇒ $\sqrt{x}=\sqrt{5–\sqrt{21}}$
⇒ $\sqrt{x}={\sqrt \frac{10–2\sqrt{21}}{2}}$
⇒ $\sqrt{x}=\sqrt{\frac{7+3–2\sqrt{7}\sqrt{3}}{2}}$
⇒ $\sqrt{x}=\sqrt{\frac{(\sqrt{7}–\sqrt{3})^{2}}{2}}$
⇒ $\sqrt{x}=\frac{\sqrt{7}–\sqrt{3}}{\sqrt{2}}$
Putting the value of $\sqrt{x}$ in the expression $\frac{\sqrt{x}}{\sqrt{32–2x}–\sqrt{21}}$, we have:
= $\frac{\frac{\sqrt{7}–\sqrt{3}}{\sqrt{2}}}{\sqrt{32–10+2\sqrt{21}}–\sqrt{21}}$
= $\frac{\frac{\sqrt{7}–\sqrt{3}}{\sqrt{2}}}{\sqrt{22+2\sqrt{21}}–\sqrt{21}}$
= $\frac{\frac{\sqrt{7}–\sqrt{3}}{\sqrt{2}}}{\sqrt{21+1+2\sqrt{21}.\sqrt{1}}–\sqrt{21}}$
= $\frac{\frac{\sqrt{7}–\sqrt{3}}{\sqrt{2}}}{\sqrt{(\sqrt{21}+\sqrt{1})^{2}}–\sqrt{21}}$
= $\frac{\frac{\sqrt{7}–\sqrt{3}}{\sqrt{2}}}{\sqrt{21}+1–\sqrt{21}}$
= $\frac{1}{\sqrt{2}}(\sqrt{7}–\sqrt{3})$
Hence, the correct answer is $\frac{1}{\sqrt{2}}(\sqrt{7}–\sqrt{3})$.
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