Solving a radical equation for real roots

Question

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 $x = -\dfrac{1}{\sqrt{2}}$.

Answer 1

$0(-1)=(-4x^4+4x^2-1)(-1) \\ 0=4x^4-4x^2+1 \\ 0=(2x^2-1)^2$

Answer 2

When you find yourself in such a situation, a good trick is to insert the root (any number will work but the root will often make the calculations easy) at each step and see when the value change.

Personally I think you are doing to much at once in the last step.

Answer 3

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.