Question

If x = 1 / 3-√5 , Then Find The Value Of √x + 1 / √x.....Pls Send The Solution Process Wise

Answer 1

i hope you will b understand this solution

Answer 2

Answer:

The value of  $\sqrt{x}+\frac{1}{\sqrt{x} } =\frac{3\sqrt{5}- 1 }{2\sqrt{2} }$

Step-by-step explanation:

Given,

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

To find,

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

Solution:

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

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

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

$= \frac{3+\sqrt{5} }{9-5}$

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

$\frac{1}{x} = \frac{4}{3+\sqrt{5} }$  

$\frac{1}{x} = \frac{4}{3+\sqrt{5} }X\frac{3-\sqrt{5} }{3-\sqrt{5} }$

$= \frac{4(3-\sqrt{5})}{9-5}}$

= 3-√5

From equation (1) we get,

$(\sqrt{x}+\frac{1}{\sqrt{x} })^2 = \frac{3+\sqrt{5} }{4}+3-\sqrt{5}+2$

$= \frac{3+\sqrt{5}+12 - 4\sqrt{5}+8 }{4}$

$= \frac{23-3\sqrt{5}}{4}$

$= \frac{2(23-3\sqrt{5})}{2X4}$

$= \frac{46 -6\sqrt{5})}{8}$

$= \frac{45 + 1 - 6\sqrt{5} }{8}$

$= \frac{(3\sqrt{5})^2 + 1^1 - 2X3\sqrt{5} }{(2\sqrt{2})^2 }$

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

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

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

The value of $\sqrt{x}+\frac{1}{\sqrt{x} } =\frac{3\sqrt{5}- 1 }{2\sqrt{2} }$

#SPJ2