Question

If $x+\frac{1}{x}=2$ then the positive value of $\sqrt{x}+\frac{1}{\sqrt{x}}$ will be

• A. $\sqrt{2}$
• B. $2$
• C. $\sqrt{2}+1$
• D. $1$

Answer 1

The correct option is B $2$

Given: $x+\frac{1}{x}=2$

$(\sqrt{x}+\frac{1}{\sqrt{x}}{)}^2=(\sqrt{x}{)}^2+(\frac{1}{\sqrt{x}}{)}^2+2\sqrt{x}.\frac{1}{\sqrt{x}}=x+\frac{1}{x}+2=2+2=4$

Hence,

$(\sqrt{x}+\frac{1}{\sqrt{x}}{)}^2$ = 4

Taking square root on both the sides we get

$\Rightarrow \sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{4}=2$