The curve of the function $f (x)$ is given.
Find the real roots of the equation $f (4-3x^2)=0.$
I know the possible solution:
From the equations $4-3x^2=-2$ and $4-3x^2=4$ we have $ x=\left\{0,\sqrt 2,- \sqrt 2\right\} $.
Now, I want to create a little proof for the alternative solution before moving on to the calculation.
It is obvious that, we don't know directly what the explicit form of $f(x)$ is.
Goal: I want to find a new $ g (4-3x ^ 2) = 0 $ equation that will definitely give the same real solutions as the equation $f (4-3x ^ 2) = 0.$
Let's define the function $g:\mathbb{R}\longrightarrow \mathbb {R}$.
For any $x\in\mathbb {R}$ if $f(x)=0$, then $g(x)=0$ and if $f(x)≠0$, then $g(x)≠0$ and there exist at least one $x=r, r\in \mathbb{R}$, which gives $f(r)≠g(r)$.
According to our definition of $g(x)$, we deduce that for any $x\in\mathbb{R}$, if $f(4-3x^2)=0,$ then $g(4-3x^2)=0$ and if $f(4-3x^2)≠0,$ then $g(4-3x^2)≠0.$
Finally, we can deduce that solving the equation $ f (4-3x ^ 2) = 0 $ is equivalent to solving the equation $ g (4-3x ^ 2) = 0. $ In other words, the real roots of the equation $ f (4-3x^2) = 0 $ are the same as the equation $ g (4-3x^2) = 0$. So, it is enough to solve the equation $g (4-3x^2) = 0$.
Let, $g(x)=(x+2)(x-4)$.
Then we have, $$g(4-3x^2)=0 \Longrightarrow (4-3x^2+2)(4-3x^2-4)=0 \Longrightarrow -3x^2(6-3x^2)=0 \Longrightarrow x=\left\{0,\sqrt 2,- \sqrt 2\right\} $$
Question: Is my proof-solution correct?