Question

If $x=3+2\sqrt{2}$ find the value of $\sqrt{x}+\frac{1}{\sqrt{x}}$

• A. $2$
• B. $2\sqrt{2}$
• C. $3$
• D. $3\sqrt{2}$

Answer 1

The correct option is B $2\sqrt{2}$

If $x=3+2\sqrt{2}$ then $\frac{1}{x}$ will be as follows:

$\frac{1}{x}=\frac{1}{3+2\sqrt{2}}=\frac{1}{3+2\sqrt{2}}\times \frac{3-2\sqrt{2}}{3-2\sqrt{2}}\left(Rationalise\right)=\frac{3-2\sqrt{2}}{3^2-(2\sqrt{2}{)}^2}(\because (a+b\left)\right(a-b)=a^2-b^2)$

$=\frac{3-2\sqrt{2}}{9-8}=\frac{3-2\sqrt{2}}{1}\Rightarrow \frac{1}{x}=3-2\sqrt{2}....\left(1\right)$

Now, consider ${(\sqrt{x}+\frac{1}{\sqrt{x}})}^2$ as shown below:

${(\sqrt{x}+\frac{1}{\sqrt{x}})}^2={\left(\sqrt{x}\right)}^2+{\left(\frac{1}{\sqrt{x}}\right)}^2+(2\times \sqrt{x}\times \frac{1}{\sqrt{x}})(\because (a+b{)}^2=a^2+b^2+2ab)=x+\frac{1}{x}+2\Rightarrow {(\sqrt{x}+\frac{1}{\sqrt{x}})}^2=x+\frac{1}{x}+2$

$\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{x+\frac{1}{x}+2}\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{3+2\sqrt{2}+3-2\sqrt{2}+2}(Given:x=3+2\sqrt{2}andfromeqn(1\left)\right)\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{8}\sqrt{x}+\frac{1}{\sqrt{x}}=2\sqrt{2}$

Hence, $\sqrt{x}+\frac{1}{\sqrt{x}}=2\sqrt{2}$.