Solving a radical equation for real roots I'm attempting to solve the derivative of my function $f(x)$ for real roots.
$$
\\ \begin{align*}
\\ f(x) &= 3x^2 + 3\arcsin{x}
\\ f^{\prime}(x) &= 6x + \dfrac{3}{\sqrt{1-x^2}}
\\ 
\\ 0 &= 6x + \dfrac{3}{\sqrt{1-x^2}}
\\
\\ 0 &= 6x \sqrt{1-x^2} + 3
\\ 0 &= \sqrt{36x^2 - 36x^4} + 3
\\ -3 &= \sqrt{36x^2-36x^4}
\\ (-3)^2 &= (\sqrt{36x^2-36x^4})^2
\\ 0 &= -36x^4 + 36x^2 - 9
\\ \frac{0}{9} &= \frac{9(-4x^4 + 4x^2 - 1)}{9}
\\ 0 &= -4x^4 + 4x^2 - 1
\\ \end{align*}
$$
I've considered the rational root theorem at this point, through which I find possible roots to be $x = \{ \pm1, \pm\frac{1}{2}, \pm\frac{1}{4} \}$.
Clearly I've made an error, as the actual root is $x = -\dfrac{1}{\sqrt{2}}$.

Your work is correct, $1/\sqrt 2$ is indeed a root of $-4x^4+4x^2-1$. The rational root test just tells you that the only possible rational roots of your polynomial are $\pm 1$, $\pm 1/2$ and $\pm 1/4$, it doesn't tell you anything about possible irrational roots, so there's no contradiction. For example, the rational root test also tells you that the only possible rational roots of $x^2-2$ are $\pm 1$ and $\pm 2$, but that doesn't contradict the fact that the actual roots are $\pm\sqrt 2$.
You can easily solve $-4x^4+4x^2-1=0$ by writing it as $-4(x^2)^2+4x^2-1=0$, solving for $x^2$ and then taking a square root.
