Question

If x=3-2root2,then find the value of root x+1/root x

Answer 1

Answer :

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

Now,

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

Now

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

Answer 2

Sry fr unclear n shabby work...

hope it helps..

mark it as brainliest