If $$x + \frac{1}{x} = \frac{{\sqrt 3 + 1}}{2},$$    then what is the value of $${x^4} + \frac{1}{{{x^4}}}?$$

Solution(By Examveda Team)

$$\eqalign{ & x + \frac{1}{x} = \frac{{\sqrt 3 + 1}}{2} \cr & {x^2} + \frac{1}{{{x^2}}} = {\left( {\frac{{\sqrt 3 + 1}}{2}} \right)^2} - 2 \cr & = \frac{{4 + 2\sqrt 3 }}{4} - 2 \cr & = \frac{{4 + 2\sqrt 3 - 8}}{4} \cr & = \frac{{ - 4 + 2\sqrt 3 }}{4} \cr & = \frac{{ - 2 + \sqrt 3 }}{2} \cr & {x^4} + \frac{1}{{{x^4}}} = {\left( {\frac{{ - 2 + \sqrt 3 }}{2}} \right)^2} - 2 \cr & = \frac{{4 + 3 - 4\sqrt 3 - 8}}{4} \cr & = \frac{{ - 1 - 4\sqrt 3 }}{4} \cr} $$