Question

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

• A. $2(3-\sqrt{2})$
• B. $6\sqrt{2}-2$
• C. $6-\sqrt{2}$
• D. 12

Answer 1

Answer: (D)

12

Since, $x=\sqrt{2+\sqrt{2}}$
$\therefore x^2=2+\sqrt{2}$
$\therefore x^4=4+2+4\sqrt{2}$
Here, $x^4=4+2+4\sqrt{2}=6+4\sqrt{2}$
then $\frac{4}{x^4}=\frac{4}{6+4\sqrt{2}}$
$=\frac{4}{6+4\sqrt{2}}\times \frac{6-4\sqrt{2}}{6-4\sqrt{2}}$
$=\frac{24-16\sqrt{2}}{36-32}$
$=\frac{24-16\sqrt{2}}{4}$
$=6-4\sqrt{2}$
$\therefore x^4+\frac{4}{x^4}=6+4\sqrt{2}+6-4\sqrt{2}=12$
Option D is correct.