Question

if
$x = \frac{1}{ \sqrt{3} + \sqrt{5} }$
find the value of
$(x + \frac{1}{x} ) {}^{2}$

Answer 1

Answer:

The numeric value of ( x + 1 / x )^2 is ( 47 + 6√10 ) / 4


Step-by-step explanation:

It is given that the value of x is $\dfrac{1}{\sqrt{3}+\sqrt{5}}$


$\implies x =\dfrac{1}{\sqrt3 + \sqrt5}$


Mutliply and divide by $\dfrac{\sqrt{3}-\sqrt{5}}{\sqrt{3}-\sqrt{5}$ on right hand side,

$\implies x= \dfrac{1}{\sqrt{3}+\sqrt5}\times \dfrac{\sqrt3-\sqrt5}{\sqrt3-\sqrt5}\\\\\\\implies x =\dfrac{\sqrt3-\sqrt5}{(\sqrt3+\sqrt5)(\sqrt3-\sqrt5)}\\\\\\\implies x =\dfrac{\sqrt3-\sqrt5}{(\sqrt3)^2-(\sqrt5)^2}\\\\\\\implies x =\dfrac{\sqrt3- \sqrt5}{3-5}=\dfrac{\sqrt3-\sqrt5}{-2}\\\\\\\implies x=\dfrac{\sqrt5-\sqrt3}{2}$

Therefore,

$\implies \dfrac{1}{x}=\dfrac{2}{\sqrt5-\sqrt3}$


Multiply and divide by $\dfrac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}$ on right hand side,

$\implies \dfrac{1}{x}=\dfrac{2}{\sqrt5-\sqrt3} \times \dfrac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}\\\\\\\implies \dfrac{1}{x}=\dfrac{2(\sqrt5+\sqrt3)}{(\sqrt5)^2-(\sqrt3)^2}=\dfrac{2(\sqrt5+\sqrt3)}{5-3}\\\\\\\implies\dfrac{1}{x}=\sqrt5+\sqrt3$



Then,

Adding x and 1 / x,

$\implies x + \dfrac{1}{x}=\dfrac{\sqrt5-\sqrt3}{2}+\sqrt5+\sqrt3\\\\\\\implies x + \dfrac{1}{x}=\dfrac{\sqrt5-\sqrt3+2\sqrt5+2\sqrt3}{2}\\\\\\\implies x+\dfrac{1}{x}=\dfrac{3\sqrt5+\sqrt2}{2}$



Square on both sides,

= > ( x + 1 / x )^2 = [( 3√5 + √2 ) / 2 ]^2

= > ( x + 1 / x )^2 = ( 45 + 2 + 6√10 ) / 4

= > ( x + 1 / x )^2 =  ( 47 + 6√10 ) / 4


Hence,

The numeric value of ( x + 1 / x )^2 is ( 47 + 6√10 ) / 4