Question

if X= 3-2root2 find the value of root X +1/root X and root x-1/root X p​

Answer 1

Given :  x = 3  - 2√2

To find : x + 1/x   & x - 1/x

Solution:

x = 3  - 2√2

1/x = 1/(3  - 2√2)

on rationalizing

= (3 + 2√2)/( 9 - 9)

= 3  + 2√2

x + 1/x  =  3  - 2√2 +  3 + 2√2

=> x + 1/x  =  6

x - 1/x  = 3  - 2√2  -(  3 + 2√2 )

=  3  - 2√2 - 3  - 2√2

= -4√2

x + 1/x  =  6

x - 1/x  = -4√2

Learn more:

if x=1/4-x, find : x+1/x - Brainly.in

https://brainly.in/question/5639594

Answer 2

Given: $x=3-2\sqrt{2}$

To find: $\sqrt{x}+\frac{1}{\sqrt{x}},\:\sqrt{x}+\frac{1}{\sqrt{x}}$

Solution:

Now, $x=3-2\sqrt{2}$

$\Rightarrow x=2-2\sqrt{2}+1$

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

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

$\Rightarrow \sqrt{x}=\sqrt{2}-1$

Then, $\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{2}-1}$

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

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

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

Now, $\sqrt{x}+\frac{1}{\sqrt{x}}$

$\quad=\sqrt{2}-1+\sqrt{2}+1$

$\quad=2\sqrt{2}$

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

and $\sqrt{x}-\frac{1}{\sqrt{x}}$

$\quad=\sqrt{2}-1-\sqrt{2}-1$

$\quad=-2$

$\Rightarrow \boxed{\sqrt{x}-\frac{1}{\sqrt{x}}=-2}$