Question

When simplified $\sqrt{1+[\frac{x^4-1}{2x^2}{]}^2}$ equals:

• A. $\frac{x^4+2x^2-1}{2x^2}$
• B. $\frac{x^4-1}{2x^2}$
• C. $\frac{\sqrt{x^2+1}}{2}$
• D. $\frac{x^2}{\sqrt{2}}$
• E. $\frac{x^2}{2}+\frac{1}{2x^2}$

Answer 1

The correct option is E $\frac{x^2}{2}+\frac{1}{2x^2}$

$\sqrt{1+{\left(\frac{x^4-1}{2x^2}\right)}^2}=\sqrt{\frac{{(x^4-1)}^2+{4x}^4}{{4x}^4}}=\sqrt{\frac{{(x^4+1)}^2}{{4x}^4}}=\frac{(x^4+1)}{2x^2}=\frac{x^2}{2}+\frac{1}{2x^2}$