Is this implication true:$ f(a)=0 \implies f'(a)=0$ with $f$ apolynomial of second degree and $\Delta=0$?

Question

Let $f(x)$ be a real valued function defined as a polynomial of second degree with $\Delta=0$. I want to know if the following is true:

If $x=a$ is a multiple root of $f(x)=0$, then $f'(a)=0$.

$f(a)=0 \implies f'(a)=0$ ?

I have checked it using many example it seems to work, but I can't seem to prove it.

Answer 1

$x=a$ is a multiple root of $f(x)$, so $(x-a)^{2}$ divides $f(x)$, say, $f(x)=(x-a)^{2}g(x)$, then $f'(x)=2(x-a)g(x)+(x-a)^{2}g'(x)$, so $f'(a)=2(a-a)g(a)+(a-a)^{2}g'(a)=0$.

Answer 2

Yes, it is true.

You are assuming that $f(x)=ax^2+bx+c$, with $b^2=4ac$ and $a\neq0$. But then $f(x)=a\left(x+\frac b{2a}\right)^2$, because$$a\left(x+\frac b{2a}\right)^2=ax^2+bx+\frac{b^2}{4a}=ax^2+bx+c=f(x).$$So, the only root is $-\frac b{2a}$ and, on the other hand, $f'(x)=2a\left(x+\frac b{2a}\right)$. So, $-\frac b{2a}$ is also a root of $f'$.