Question

if x is equals to 9 + 4 root 5 find the value of root x minus one by root x

Answer 1

Given x = 9 + 4√5.

⇒ 4 + 5 + 2 * 2 * √5

⇒ (2)^2 + (√5)^2 + 2 * (2) * (√5)

It is in the form of a^2 + b^2 + 2ab = (a + b)^2.

⇒ (2 + √5)^2.

So, x = (2 + √5)^2.

Then, √x = (2 + √5).

Now,

$= > \frac{1}{\sqrt{x}}$

$= > \frac{1}{(2 + \sqrt{5})} * \frac{(2 - \sqrt{5})}{(2 - \sqrt{5})}$

$= > \frac{(2 - \sqrt{5})}{(2)^2 - (\sqrt{5})^2}$

$= > \frac{2 - \sqrt{5}}{4 - 5}$

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

$= > \sqrt{5} - 2$

Hence,

⇒ √x - (1/√x)

⇒ (2 + √5) - (√5 - 2)

⇒ 2 + √5 - √5 - 2

⇒ 4.

Therefore, the value of √x - (1/√x) = 4.

Hope this helps!

Answer 2

"$\sqrt { x } - \frac { 1 } { \sqrt { x } } = 4$

Given:

$x = 9 + 4 \sqrt { 5 }$

To find:

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

Solution:

$x = 9 + 4 \sqrt { 5 }$

$= 4 + 5 + 2 \times 2 \times \sqrt { 5 }$

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

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

$\therefore \sqrt { x } = 2 + \sqrt { 5 } (neglecting the negative sign)$

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

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

$= \frac { 2 - \sqrt { 5 } } { \left\{ ( 2 ) ^ { 2 } - ( \sqrt { 5 } ) ^ { 2 } \right\} }$

$= \frac { 2 - \sqrt { 5 } } { 4 - 5 }$

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

$= \sqrt { 5 } - 2$

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

$= 2 + \sqrt { 5 } - \sqrt { 5 } + 2$

$\sqrt { x } - \frac { 1 } { \sqrt { x } } = 4$"