Question

If $x=\sqrt{2+\sqrt{\sqrt{2}}},$ then $\frac{4}{x^4}$ is :

• A. $\frac{4}{4+\sqrt{2}}$
• B. $6\sqrt{2}-2$
• C. $6\sqrt{2}+2$
• D. $12$

Answer 1

Answer: (A)

$\frac{4}{4+\sqrt{2}}$

$\begin{array}{c}wehave\\ x=\sqrt{2+\sqrt{\sqrt{2}}}\\ Now,\\ \frac{4}{x^4}=\frac{4}{{\left(\sqrt{2+\sqrt{\sqrt{2}}}\right)}^4}\\ =\frac{4}{{\left(\sqrt{2+\sqrt{{\left(2\right)}^{\frac{1}{2}}}}\right)}^4}\\ =\frac{4}{{\left(\sqrt{2+{\left(2\right)}^{\frac{1}{4}}}\right)}^4}\\ =\frac{4}{{\left(2+2^{\frac{1}{4}}\right)}^{\frac{1}{2}\times 4}}\\ =\frac{4}{2^2+{\left(2\right)}^{\frac{1}{4}\times 2}}\\ =\frac{4}{4+{\left(2\right)}^{\frac{1}{2}}}\\ =\frac{4}{4+\sqrt{2}}\\ Hence,theoptionAisthecorrectanswer.\end{array}$