If $${x^2} + \frac{1}{{{x^2}}} = \frac{7}{4}$$   for x > 0 then what is the value of $${x^4} + \frac{1}{{{x^4}}}.$$

A. $$1$$

B. $$\frac{{17}}{{16}}$$

C. $$\frac{{15}}{{16}}$$

D. $$\frac{{51}}{{16}}$$

Answer: Option B

Solution(By Examveda Team)

$$\eqalign{ & {\text{Given, }}{x^2} + \frac{1}{{{x^2}}} = \frac{7}{4} \cr & {\text{Squaring both sides, we get}} \cr & {x^4} + \frac{1}{{{x^4}}} + 2 = \frac{{49}}{{16}} \cr & {x^4} + \frac{1}{{{x^4}}} = \frac{{49}}{{16}} - 2 \cr & {x^4} + \frac{1}{{{x^4}}} = \frac{{49 - 32}}{{16}} \cr & {x^4} + \frac{1}{{{x^4}}} = \frac{{17}}{{16}} \cr} $$