Solutions

a) If $f=g^2$ for some $g$, then $f$ must be positive since $g^2$ is positive for any $g$.

b) Notice that

$$g=\frac{1}{f}$$

so $f$ cannot be equal to $0$. Thus, any function whose range excludes $0$ is the answer.

c) Using the quadratic formula, the zeros of $x(t)$ are

$$x_0=\frac{-b(t)\pm\sqrt{b(t)^2-4c(t)}}{2}.$$

So the condition on $b(t)$ and $c(t)$ is that they must satisfy

$$b(t)^2-4c(t)\ge 0 \implies b(t)^2\ge 4c(t)$$

for all numbers $t$. We can then choose $x_0$ as the function $x$ such that the equation holds for all $t$.

d) Solving for $x(t)$, we get

$$x(t)=\frac{-b(t)}{a(t)}.$$

In the equation above, $a(t)$ cannot equal $0$. However, if it does, the original equation is trivially satisfied with

$$a(t)=0$$

and

$$b(t)=0,$$

in which case $x(t)$ has infinitely many solutions, as it can be any function. On the other hand, if the equation we get solving for $x(t)$ does hold, then $a(t)$ cannot be zero and the only solution for $x(t)$ is that equation.