Will the roots change

$$x_{1/2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

If you are looking at the limit for $a\to 0$ then the equation becomes a linear equation $bx+c=0$.

The roots are giving (on any field of characteristic $\neq2$) by:

$$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Taking this as a function of $\;a\;$ , we can apply l'Hospital's rule for one of the roots:

$$\lim_{a\to0}\frac{-b+\sqrt{b^2-4ac}}{2a}=\lim_{a\to0}-\frac{c}{\sqrt{b^2-4ac}}=-\frac cb$$

assuming, of course, $\;b\neq0\;$ .

For the other one the numerator dfoesn't vanish but the denominator does so the limit is $\;\pm\infty\;$ , depending on the sign of $\;b\;$ .

If $|a|$ is small enough, the discriminant $b^2-4ac$ will be positive, so the equation can be considered to have two distinct roots, by keeping $a\in(-\delta,\delta)$, $a\ne0$, for some $\delta>0$.

We can assume $b>0$ (just change $x$ to $-x$ otherwise)

The roots are $$ \frac{-b-\sqrt{b^2-4ac}}{2a}, \qquad \frac{-b+\sqrt{b^2-4ac}}{2a}, $$ We see that $$ \lim_{a\to0}\frac{\sqrt{b^2-4ac}-b}{2a}= \lim_{a\to0}\frac{-4ac}{2a(\sqrt{b^2-4ac}+b)}=-\frac{c}{b} $$ whereas $$ \lim_{a\to0^+}\frac{\sqrt{b^2-4ac}+b}{2a}=-\infty \qquad \lim_{a\to0^-}\frac{\sqrt{b^2-4ac}+b}{2a}=\infty $$

In my old high school book, when discussing “parametric degree two equations” it was always said that “when $a=0$ one root becomes infinite”, which has a justification: the parametric equation was essentially a pencil of parabolas, the degenerate ones being a line counted twice (case of discriminant $0$) and the union of a line and of the improper line (case $a=0$). Of course the book didn't mention this.