When simplified √1+((x^4-1)/(2x^2))^2 equals:

1954 AMC 12 Problem 14

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14.

When simplified

\sqrt{1+\left(\dfrac{x^4-1}{2x^2}\right)^2}

equals:

\dfrac{x^4+2x^2-1}{2x^2}

\dfrac{x^4-1}{2x^2}

\dfrac{\sqrt{x^2+1}}2

\dfrac{x^2}{\sqrt2}

\dfrac{x^2}{2}+\dfrac1{2x^2}

Put

1

over the denominator

4x^4

The numerator becomes

4x^4+(x^4-1)^2=(x^4+1)^2

For

x\ne0,

\begin{aligned} &1+\left(\frac{x^4-1}{2x^2}\right)^2\\ &\quad=\frac{4x^4+(x^4-1)^2}{4x^4}\\ &\quad=\frac{(x^4+1)^2}{4x^4}. \end{aligned}

Because

x^2\gt0,

its square root is

\frac{x^4+1}{2x^2} =\frac{x^2}{2}+\frac1{2x^2}.

Thus, the correct answer is E.