Question

If $x=3+2\sqrt{2}$, then the value of $(\sqrt{x}-\frac{1}{\sqrt{x}})$ = .

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

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

Given, $x=3+2\sqrt{2}$

Substituting the value of $x$ in $x+\frac{1}{x}-2$,

$(3+2\sqrt{2})+\frac{1}{(3+2\sqrt{2})}-2$

= $3+2\sqrt{2}+\frac{3-2\sqrt{2}}{(3+2\sqrt{2})(3-2\sqrt{2})}-2$

= $3+2\sqrt{2}+\frac{3-2\sqrt{2}}{9-8}-2$

= $3+2\sqrt{2}+\frac{3-2\sqrt{2}}{1}-2$

= $3+2\sqrt{2}+3-2\sqrt{2}-2$

=$6-2$

= $4$
Since, ${(\sqrt{x}-\frac{1}{\sqrt{x}})}^2=4,(\sqrt{x}-\frac{1}{\sqrt{x}})=2$ .