Let $a = (a_0, \dots, a_{n-1})^T \in \mathbb R^n$ and $b = (b_0, \dots, b_{n-1})^T \in \mathbb R^n$. We define two polynomials by
\begin{align*}
f_a(x) &= x^n + a_{n-1} x^{n-1} + \cdots + a_0 \\
f_b(x) &= b_{n-1} x^{n-1} + \cdots +b_0.
\end{align*}
Suppose we require $f_a$ and $f_b$ are coprime in $\mathbb C$. Can we find a scalar $s \in \mathbb R$ such that $f_a(x) + s f_b(x)$ has distinct roots in $\mathbb C$?
If $f_a(x) + s f_b(x) = x^n + (a_{n-1} + s b_{n-1})x^{n-1} + \cdots + (a_0 + sb_0)$ has multiple roots for all $s \in \mathbb R$, this would imply the discriminant which is a polynomial in $s$ would be identically $0$ for $s \in \mathbb R$. I don't think this can happen but could not figure out how to argue this part.
To not cause confusion, this sentence is crossed out. This should be a separate question.
The coprime condition just conveniently comes out of my situation. I am not sure this is necessary. Intuitively, I would think $b \neq 0$ should enough.
Best Answer
Suppose that $f(x)+sg(x)$ has a repeated root for all $s$. Then for each $s$ there exists $x_s$ such that $$f(x_s)+sg(x_s)=f'(x_s)+sg'(x_s)=0.$$ Therefore, $f(x_s)g'(x_s)-f'(x_s)g(x_s)=0$ for all $s$. Let $h(x)=f(x)g'(x)-f'(x)g(x)$ and let $A$ be the finite set of roots of $h(x)$. $A$ is finite because otherwise $h(x)=0$ which implies that $f(x)g'(x)=f'(x)g(x)$. Since $f(x)$ and $g(x)$ are coprime, this in turn implies that $f(x)\mid f'(x)$ which is not possible.
It follows that $x_s \in A$ for all $s$ and so there exists $x_0$ such that $x_s=x_0$ for infinitely many $s$. One has $f(x_0)+sg(x_0)=0$ for infinitely many values of $s$. This clearly happens only if $f(x_0)=g(x_0)=0$ contradicting the assumption that $f(x),g(x)$ are coprime.