Question

If x=1/3-root5 then what is the value of Rootx-1/rootx

Answer 1

Step-by-step explanation:

Given:

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

To Find :

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

Solution:

Let's first rationalise the value of x,

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

$x = \frac{3 + \sqrt{5} }{4}$

(Use following identity for denominator: (a+b)(a-b) = a²-b² )

Now,

Let's find √x.

$\sqrt{x} = \sqrt{ \frac{3 + \sqrt{5} }{4} }$

$= \frac{ \sqrt{3 + \sqrt{5} } }{ \sqrt{4} }$

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

Now,

We have to find,

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

Substituting the value of √x,

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

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

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

$= \frac{3 + \sqrt{5} - 4 }{2 \sqrt{3 + \sqrt{5} } }$

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

Again rationalising the denominator,

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

$= \frac{( \sqrt{5} - 1)( \sqrt{3 + \sqrt{5}) } }{6 + 2 \sqrt{5} }$

Again rationalising the denominator : )

$\frac{( \sqrt{5} - 1)( \sqrt{3 + \sqrt{5} } )(6 - 2 \sqrt{5)} }{16}$

I hope I am correct !