Question : If $x=3 - 2\sqrt2$, then $\sqrt x+(\frac{1}{\sqrt x})$ is equal to:
Option 1: $0$
Option 2: $1$
Option 3: $2$
Option 4: $2\sqrt2$
Correct Answer: $2\sqrt2$
Solution :
Given:
$x = 3 - 2\sqrt{2}$
According to the question,
Let $y = \sqrt{x}+\frac{1}{\sqrt{x}}$
Squaring both sides,
$y^2 = x + \frac{1}{x}+2$
Put the value $x$,
$y^2 = 3 - 2\sqrt{2} + \frac{1}{3-2\sqrt{2}} + 2$
or, $y^2 = 3 - 2\sqrt{2} + \frac{1}{3-2\sqrt{2}} × \frac{3+\sqrt{2}}{3 + \sqrt{2}} + 2$
or, $y^2 = 5 - 2\sqrt{2} + 3 + 2\sqrt{2}$
or, $y^2 = 8$
or, $y = 2\sqrt{2}$
Hence, the correct answer is $2\sqrt{2}$.
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