Problem: Given two polynomials of equal degree:
\begin{equation} p_1(x)=x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_0,\\ p_2(x)=x^n+(a_1+\epsilon)x^{n-1}+a_2x^{n-2}+\cdots+a_0. \end{equation}
where $\epsilon\geq0.$
What are conditions on $\epsilon$ and $a_i's$, such that roots of $p_2(x)$ located to the left of roots $p_1(x)$? My aim is to move roots of $p_1(x)$ slightly to the left using the $\epsilon$. For example, if roots of $p_1(x)$ are $-2$ and $4$, I want the roots of $p_2(x)$ to be $-2.1$ and $3.8$.
My attempt:
1) Lets first consider case when $n=1$, then condition is $\epsilon>0$.
2) Now let $n=2$. Then roots of $p_1(x)$ are $\frac{-a_1-\sqrt{a_1^2-4a_2}}{2}$ and $\frac{-a_1+\sqrt{a_1^2-4a_2}}{2}$,
$\qquad\qquad$ while roots of $p_2(x)$ are $\frac{-a_1-\epsilon-\sqrt{a_1^2+2a_1\epsilon+\epsilon^2-4a_2}}{2}$ and $\frac{-a_1-\epsilon+\sqrt{a_1^2+2a_1\epsilon+\epsilon^2-4a_2}}{2}$.
We want the difference of first roots to be positive and difference of second roots to be positive as well, i.e. simultaneously satisfy the following two inequalities:
\begin{equation} \epsilon+\sqrt{a_1^2+2a_1\epsilon+\epsilon^2-4a_2}-\sqrt{a_1^2-4a_2}>0, \qquad\qquad\qquad (1)\\ \epsilon-\sqrt{a_1^2+2a_1\epsilon+\epsilon^2-4a_2}+\sqrt{a_1^2-4a_2}>0. \qquad\qquad\qquad (2) \end{equation}
In $(1)$ and $(2)$ I assume that square root parts are not complex. Then after some calculation, I found that conditions are $a_1<0$, $a_2<0$ and $\epsilon>\sqrt{a_1^2-4a_2}$.
Is it possible to generalize to larger degrees? Is there any intuition on geometry of roots when small perturbations are made to $a_1$ or equal amount of small perturbations are made to all $a_i's$?