Question

plz answer fast its urgent

Answer 1

hey mate I hope my answer helps you...

according to the question x = 7+√40,
then √x = √[7+√40]
Here the value of x is a surd then the conjugate for it is √[7-√40]

so √x+1/√x = √[7+√40]+√[7-√40]
therefore we get √7+7+√40-√40
so we get 14 as the answer

pls mark it as brainliest..... I wish you are satisfied with my answer

Answer 2

Given ---
x = 7 + √40

To find ---
√x + 1/√x

Finding ---

$x = 7 + \sqrt{40}$

$= > \sqrt{x} = \sqrt{7 + \sqrt{40} }$

Now, √40 =
√(2×2×2×5)

=> √40 = 2×√10

=> √40 = 2√10

$= > \sqrt{x} = \sqrt{7 + 2 \sqrt{10} }$

$= > \sqrt{x} = \sqrt{5 + 2 + 2 \sqrt{10} }$

$\sqrt{x} = \sqrt{ ({ \sqrt{5} )}^{2} + ({ \sqrt{2} )}^{2} + (2 \times \sqrt{2} \times \sqrt{5} ) }$

Now, using the formula
(A+B)² = A²+B²+AB, we get,

$= > \sqrt{x} = \sqrt{( { \sqrt{5 } + \sqrt{2} )}^{2} }$

Cancelling the ^2 and √, we get,

$= > \sqrt{x} = \sqrt{5} + \sqrt{2}$

Hence, √x = √5+√2......(equation 1)

Now,
$\frac{1}{ \sqrt{x} } = \frac{1}{ \sqrt{5} + \sqrt{2} }$

Rationalising the denominator, we get,

$= > \frac{1}{ \sqrt{x} } = \frac{1}{ \sqrt{5} + \sqrt{2} } \times \frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} - \sqrt{2} }$

$= > \frac{1}{ \sqrt{x} } = \frac{1 \times \sqrt{5} - \sqrt{2} }{ { \sqrt{5} }^{2} - { \sqrt{2} }^{2} }$

$= > \frac{1}{ \sqrt{x} } = \frac{ \sqrt{5} - \sqrt{2} }{5 - 2}$

$= > \frac{1}{ \sqrt{x} } = \frac{ \sqrt{5} - \sqrt{2} }{3}$

Hence, 1/√x = (√5-√2)/3......(equation 2)

Now, from equations 1 and 2

$\sqrt{x} + \frac{1}{ \sqrt{x} } = ( \sqrt{5} + \sqrt{2} ) + ( \frac{ \sqrt{5} - \sqrt{2} }{3} )$

$= > \sqrt{x} + \frac{1}{ \sqrt{x} } = \sqrt{5} + \sqrt{2} + \frac{ \sqrt{5} - \sqrt{2} }{3}$

$= > \sqrt{x} + \frac{1}{ \sqrt{x} } = \frac{ \sqrt{5} + \sqrt{2} + \sqrt{5} - \sqrt{2} }{3}$

Cancelling √2 and -√2, we get

$\sqrt{x} + \frac{1}{ \sqrt{x} } = \frac{2 \sqrt{5} }{3}$

Hence, √x + 1/√x = 2√5 / 3.
That's your answer.

Hope it'll help.. :-D