Solutions

a) In the forward direction $(\Rightarrow)$, suppose that $a$ is a double root of $f$. Then $a$ is a root of $f'$ since $f(x)=(x-a)^2g(x)$. Furthermore,

$$f'(x)=2(x-a)g(x)+(x-a)^2g'(x).$$

So $f'(a)=0$, and therefore $a$ is also a root of $f'$.

In the backward direction $(\Leftarrow)$, suppose that $a$ is a root of both $f$ and $f'$. We can therefore write $f(x)=(x-a)q(x)$ for some polynomial function $q$. Differentiating this function gives

$$f'(x)=q(x)+(x-a)q'(x).$$

But since $a$ is a root of $f'$, $q(a)$ must have $a$ as a root. Hence, $q$ can be written as $q(x)=(x-a)h(x)$ for some polynomial function $h$, and thus $f(x)=(x-a)^2h(x)$, thereby proving that $a$ is a double root of $f$.

b) Notice that $f(x)=0$ when

$$ax^2+bx+c=0\implies x^2+\frac{b}{a}x+\frac{c}{a}=0.$$

Then

$$\begin{align*} &\implies x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2+\frac{c}{a}=0 \\ &\implies \left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2} \\ &\implies x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a} \\ &\implies x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. \end{align*}$$

Hence, for a double root to occur, we need $b^2=4ac$. Geometrically, it says that $f(x)=0$ only when

$$x=\frac{-b}{2a}.$$