The problem requests I "eliminate the parameter to find a Cartesian equation of the curve."
The given equations are:
$x = t^2 $
and $y = t^5 $
I wasn't having trouble with these problems until this one, where when attempting to isolate $t$ in the $x$ equation, you get two different equations:
$ t = -(x)^{\frac 12} $ and $ t = (x)^{\frac 12} $
How do I go about accounting for both eqations when substituting my $t$ in the y equation $y = t^5 $?
Scrutiny of the two parametric equations
$x = t^2, \tag 1$
$y = t^5, \tag 2$
indicates that, if we assume the parameter $t \in \Bbb R$ with no other restrictions, so that $t$ is allowed to vary over all of $\Bbb R$, $y$ will also take any value in $\Bbb R = (-\infty, \infty)$, whereas $x$ is restricted to lie in the range $[0, \infty)$; this is of course intuitively seen to be the case since odd powers of $t$ preserve its sign, whereas even powers "fold" the $t$-axis back on itself at $t = 0$, so that both negative and positive values of $t$ map to the positives in $\Bbb R$; $0$ of course maps to itself for both even and odd exponents.
We note that, algebraically, we have
$x^5 = (t^2)^5 = t^{10}, \tag 3$
and
$y^2 = (t^5)^2 = t^{10}, \tag 4$
so that a necessary condition for (1), (2) is
$y^2 = x^5, \tag 5$
or
$y^2 - x^5 = 0; \tag 6$
every point $(x, y)$ satisfying (1), (2) also obeys (6).
Going the other way, that is, trying to work backwards from (5)-(6) towards (1)-(2), we would like to find, for any pair $(x, y)$ obeying (5), (6), some $t \in \Bbb R$ such that (1)-(2) bind. We observe that (5)-(6) restrict $x \in [0, \infty)$, but allow $y$ to take on any value in $\Bbb R$; therefore we may always find $t \in \Bbb R$ such that (2) holds; furthermore, such a $t$ is uniquely determined by $y$, since the map
$t \mapsto t^5 \tag 7$
is a differentiable bijection from $\Bbb R$ to itself; accepting then that $y = t^5$, (5) yields
$x^5 = y^2 = t^{10} \ge 0; \tag 8$
we may then apply the function
$\sqrt[5]{\cdot}: \Bbb R_{\ge 0} \to \Bbb R_{\ge 0}, \tag 9$
to each side of (8) to find
$x = \sqrt[5]{x^5} = \sqrt[5]{t^{10}} = t^2, \tag{10}$
re-creating the original parametric form (1)-(2). We thus see that (5)-(6) are also sufficient for (1)-(2); thus the Cartesian forms (5)-(6) are equivalent to the parametric presentation (1)-(2).