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UPSC CDS 1 2022 Math Paper

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Question : 35 of 100

Marks: +1, -0

If

x=7+4\sqrt{\overline{3}}

, then what is the value of

\sqrt{\overline{x}}+\frac{1}{\sqrt{x}}\;?

1
2
3
4

Solution: 👈: Video Solution

Given:

x=7+4\sqrt{3}

Formula Used:

a^{2}+b^{2}+2ab=(a+b)^{2}

Calculation:

We have

x=7+4\sqrt{3}

\Rightarrow x=4+3+2\cdot2\cdot\sqrt{3}

\Rightarrow x=(2)^{2}+(\sqrt{3})^{2}+2\cdot2\cdot\sqrt{3}

\Rightarrow x=(2+\sqrt{3})^{2}

\Rightarrow \sqrt{x}=2+\sqrt{3}\;\;\cdots-(i)

So,

\frac{1}{\sqrt{x}}=\frac{1}{2+\sqrt{3}}

After rationalizing we get

\frac{1}{\sqrt{\overline{x}}}=2-\sqrt{\overline{3}}

On adding equation (i) and (ii), we get

\sqrt{\overline{x}}+\frac{1}{\sqrt{x}}=2+\sqrt{3}+2-\sqrt{3}

\therefore \sqrt{\overline{x}}+\frac{1}{\sqrt{x}}=4\;\text{". "}\;

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