Question

if x=4+2√3 then find the value of √x+1/√x

Answer 1

Solution:-

given 》 x=4+2√3


$\sqrt{x} + \frac{1}{ \sqrt{x} } \\ \sqrt{4 + 2 \sqrt{3} } + \frac{1}{ \sqrt{4 + 2 \sqrt{3} } } \\ \frac{4 + 2 \sqrt{3} + 1}{ \sqrt{4 + 2 \sqrt{3} } } ans$

Answer 2

The value of √x + (1/√x) = (3√3+1)/2

⇒ $\sqrt{x} +\frac{1}{\sqrt{x} } = \frac{3\sqrt{3}+1 }{2}$  

Given:

x = 4+2√3

To find:

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

Solution:

Given that  x = 4+2√3  

x = 1 + 3 + 2√3       [ split 4 as 3 + 1 ]

x = (1)²+ (√3)² + (1)(2) (√3)

x = [ 1 + √3]²                            [ from (a+b)² = a²+ b²+2ab ]

√x = (1+√3) ___ (1)  

From (1)

$\frac{1}{\sqrt{x} } = \frac{1}{1 + \sqrt{3} }$        

To rationalize $\frac{1}{1 + \sqrt{3} }$  divide and multiply with (1 - √3)  

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

$\frac{1}{\sqrt{x} } = \frac{1-\sqrt{3} }{(1)^{2} -(\sqrt{3})^{2} }$              [ from (a+b)(a-b) = (a²-b²) ]

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

$\frac{1}{ \sqrt{x} } = \frac{1 - \sqrt{3} }{-2}$  ___ (2)

From (1) and (2)

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

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

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

=  $\frac{-1- 3\sqrt{3} }{-2}$

=  $\frac{1+ 3\sqrt{3} }{2}$  

⇒ $\sqrt{x} +\frac{1}{\sqrt{x} } = \frac{3\sqrt{3}+1 }{2}$  

The value of √x + (1/√x) = (3√3+1)/2

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