Question

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

Answer 1

Hi ,

It is given that ,

x = 3 + √2

√x = √ ( 3 + √2 )

= √ [ ( 2 + 1 ) + 2 ( 2 × 1 )

√x = √2 + 1 ---- ( 1 )

1/√x = 1/ ( √2 + 1 )

= ( √2 - 1 ) / [ ( √2 + 1 ) ( √2 - 1 ) ]

= ( √2 - 1 ) / [ ( √2 )² - 1² ]

= ( √2 - 1 ) / ( 2 - 1 )

1/√x = √2 - 1 ----( 2 )

a ) value of ( √x + 1/√x )

= ( 1 ) + ( 2 )

= √2 + 1 + √2 - 1

= 2√2

b ) value of ( √x - 1/√x )

= ( √2 + 1 ) - ( √2 - 1 )

= √2 + 1 - √2 + 1

= 2

I hope this helps you.

:)

Answer 2

Heya User, 

=_=

--> x = 3 + 2√2
               = (√2)² + 2(√2) + 1
               = [ √2 + 1 ]²
--> √x = ( √2 + 1 )

Also, -->
----> ( 1 / √x )
                    = 1 / ( √2 + 1 )
                    = ( √2 - 1 ) / ( 2 - 1 ) = ( √2 - 1 )

Hence, √x + ( 1/√x ) = [ √2 + 1 + √2 - 1 ] = 2√2
And,     √x -  ( 1/√x ) = [ √2 + 1 - √2 + 1 ] = 2 

^_^ There there.... We're done