If x⁴ + x^-4 = 194, x > 0, then what is the value of $$x + \frac{1}{x} + 2?$$

Solution (By Examveda Team)

$$\eqalign{ & {x^4} + {x^{ - 4}} = 194 \cr & {\left( {{x^2} + \frac{1}{{{x^2}}}} \right)^2} = 194 + 2 \cr & {\left( {{x^2} + \frac{1}{{{x^2}}}} \right)^2} = 196 \cr & {x^2} + \frac{1}{{{x^2}}} = 14 \cr & x + \frac{1}{x} = {\left( {16} \right)^{\frac{1}{2}}} \cr & x + \frac{1}{x} = 4 \cr & x + \frac{1}{x} + 2 = 4 + 2 \cr & x + \frac{1}{x} = 6 \cr} $$