Question

if x = 3 - 2√2, find the value of √x+1/√x

Answer 1

Answer in attachment: Please comment if you know it's wrong.

Answer 2

Answer:

The value of the expression is $\sqrt{x}+\frac{1}{\sqrt{x}}=2\sqrt2$      

Step-by-step explanation:

Given : $x=3-2\sqrt2$

To find : The value of $\sqrt{x}+\frac{1}{\sqrt{x}}$

Solution :

First we find, the value of $\frac{1}{x}$

$\frac{1}{x}=\frac{1}{3-2\sqrt2}$

Rationalize,

$\frac{1}{x}=\frac{1}{3-2\sqrt2}\times \frac{3+2\sqrt2}{3+2\sqrt2}$

$\frac{1}{x}=\frac{3+2\sqrt2}{3^2-(2\sqrt2)^2}$

$\frac{1}{x}=\frac{3+2\sqrt2}{9-8}$

$\frac{1}{x}=3+2\sqrt2$

Now,

Expression $\sqrt{x}+\frac{1}{\sqrt{x}}$

Squaring the expression,

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

Substitute the values,

$(\sqrt{x}+\frac{1}{\sqrt{x}})^2=3-2\sqrt2+3+2\sqrt2+2$

$(\sqrt{x}+\frac{1}{\sqrt{x}})^2=6+2$

$(\sqrt{x}+\frac{1}{\sqrt{x}})^2=8$

Taking root both side,

$\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt8$

$\sqrt{x}+\frac{1}{\sqrt{x}}=2\sqrt2$

Therefore, The value of the expression is $\sqrt{x}+\frac{1}{\sqrt{x}}=2\sqrt2$