The symbol $|a|$ means $a$ is a positive number or zero, and $-a$ if $a$ is a negative number. 
For all real values of $t$ the expression $\sqrt{t^4+t^2}$ is equal to?
$\textbf{(A)}\ t^3\qquad \textbf{(B)}\ t^2+t\qquad \textbf{(C)}\ |t^2+t|\qquad \textbf{(D)}\ t\sqrt{t^2+1}\qquad \textbf{(E)}\ |t|\sqrt{1+t^2}$
Factor out the $t^2$ inside the square root.
\[\sqrt{t^2 \cdot (t^2 + 1)}\]
\[\sqrt{t^2} \cdot \sqrt{t^2 + 1}\]
Remember that $\sqrt{t^2} = |t|$ because square rooting a nonnegative real number will always result in a [nonnegative number](https://artofproblemsolving.com/wiki/index.php/Nonnegative_number).
\[|t|\sqrt{t^2 + 1}\]
The answer is $\boxed{\textbf{(E)}}$.