Square root vs raising to $\frac{1}{2}$ What about the +/- ??? These seem equivalent, yet the raising to 1/2 seems to ignore the +/- aspect of a square root.   Is one more valid than the other?

 A: In both problems you are taking the square root of both sides, in the first one you're just using exponents so the +/- is still required
A: $v^2=8t$
$\iff v^2-8t=0$
$\iff  (v+\sqrt{8t})(v-\sqrt{8t})=0$
$\iff v=\pm \sqrt{8t}$.
A: Both are the same. But note that $(v^2)^{1/2}=|v|,$ since $v$ can be negative. And thus, we have $$|v|=(v^2)^{1/2}=(8t)^{1/2}\iff v=\pm \sqrt{8t}.$$
A: Short answer: you must still consider the double solution $\pm (8t)^{\frac12}$.
Long answer: Whenever you deal with rational powers of real numbers, you must be mindful of every step you go through.
That is because every time you use a rational power with an even denominator, a square root is "hidden" in that power, which means that the operation of raising to that power is only defined for positive arguments, and that there really are two answers, represented with the $\pm x$ notation.
It is easier to understand where this comes from by remembering that the equation $y=4^{\frac12}$ is equivalent to $y^2=4$, which is really just asking "Which number(s), squared, total 4?"
And the answer is obviously both $2$ and $-2$.
Similarly, whenever we write $y=4^{\frac32}$, we know it is equivalent to writing $y^2=4^3$, or asking "Which number(s), squared, is equal to the cube of 4?"
Again, the answer is both $8$ and $-8$.
However, if we were to write $y = (-4)^\frac12$, then this expression, being equivalent to $y^2 = -4$, would be like asking "What number(s), squared, equal $-4$?
This time, the answer is obviously "no (real) numbers": rational powers and roots really are the same thing, and thus obey the same rules, even if they look different.
This phenomenon happens every time you use a rational power with even denominator, and must be accounted for.
A: When you get a little further in math, you'll be told that raising a number to any exponent with a non-integer absolute value is what's called a "multi-valued function."  It's just that with a rational exponent with an even denominator it's two of these that are real rather than one as for any other real exponent.  Most of the time, the "principal value" is used, which for a positive real parameter is the positive real value.
