What will happen to the roots of the quadratic equation $$ax^2 + bx + c = 0$$ if the coefficient $a$ approaches zero while the coefficients $b$ and $c$ are constant, and $b \neq 0$?
$\lim\limits_{a \to 0}{(ax^2 + bx + c)} = bx + c = 0 \longrightarrow x = -\frac{c}{b}$
However, I don't think my solution is complete; shouldn't I end up with $2$ roots? (I've only found $1$.)
Is there indeed another root to find? If so, how to I find it?
Quadratic equation $ax^x+bx+c=0$ has two roots, $x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$. We can investigate their behavior when $a \to 0$ by calculating their limits. We assume $b>0$ (we can always mupltiply the equation by -1): \begin{split} \lim_{a\to 0}\frac{-b+\sqrt{b^2-4ac}}{2a}=&\lim_{a\to 0}\frac{-b+\sqrt{b^2-4ac}}{2a} \cdot\frac{\sqrt{b^2-4ac}+b}{\sqrt{b^2-4ac}+b} \\ = & \lim_{a\to 0}\frac{b^2-4ac-b^2}{2a(\sqrt{b^2-4ac}+b)} \\ = & \lim_{a \to 0}\frac{-4ac}{2a(\sqrt{b^2-4ac}+b)} \\ = & \lim_{a \to 0}\frac{-4c}{2(\sqrt{b^2-4ac}+b)} \\ = & -\frac{c}{b}, \\ \lim_{a\to 0}\frac{-b-\sqrt{b^2-4ac}}{2a}=&\lim_{a\to 0}\frac{-2b}{2a} \\ = &\lim_{a\to 0}\left( -\frac{b}{a}\right). \end{split} The last expression doesn't only depend on the sign of $b$ but also on the sign of $a$, i.e. the direction from which we're approaching zero, so the limit does not exist. The one-sided limits are equal to $\pm \infty$.
According to Vieta's formulas
$$x_{1}+x_{2}=-{\frac {b}{a}},\quad x_{1}x_{2}={\frac {c}{a}},$$ for the roots $x_1,x_2$. When $a$ approaches zero, both these quantities become infinite.
For a small $a$ your polinomial will be very small near the point $x = -\frac{b}c$, so one root would be around there. The other will tend to infinity according to Vieta's formulas.
The roots are $f(a)=x_1=^{{-b+\sqrt{b^2-4ac}}\over{2a}}=$ and =$g(a)=x_2=^{{-b-\sqrt{b^2-4ac}}\over{2a}}$. If the limit of $a$ is $0$ and $b>0 $, the limit of $g(a)$ is $-\infty$ if $a$ converges towards $0^+$ and is $+\infty$ if $a$ converges towards $0^-$. $f(a)$ is not determined, so you can apply Hospital and the limit is $-{c\over b}$. If $<0$ you obtain a similar result mutatis mutandis.
It's probably easiest to look at the equation as $$ ax^2 = -bx-c $$ Then the right-hand side is the same linear function as $a$ varies. The left side is a parabola with apex $(0,0)$ that flattens out towards the $x$-axis as $a$ approaches $0$. What happens to the roots depends on $b$ and $c$:
If $b=0$, then the two roots diverge towards $\pm\infty$, one in each direction.
(In the degenerate case $b=c=0$, the double root at $x=0$ stays put as long as $a$ is nonzero, of course).
Otherwise, one of the roots will converge towards the $x$-intercept of of the line, and the other one diverges towards $\pm\infty$, with sign depending on whether $a$ approaches $0$ from above or below.
If $a = 0$ then $ax^2 + bx + c = bx + c$ and the equation $bx +c=0$ is a linear equation with a single root;$x = -\frac cb$.
The set of equations $\alpha x^2 + bx^2 + c = 0$ will each have two roots of $\frac {- b \pm \sqrt {b^2 - 4\alpha c}}{2\alpha}$
We can solve $\lim_{\alpha \rightarrow 0}\frac {- b \pm \sqrt {b^2 - 4\alpha c}}{2\alpha}$ with L'Hopital.
$\lim \frac {- b \pm \sqrt {b^2 - 4\alpha c}}{2\alpha}= \frac{\mp 2c\frac 1{\sqrt{b^2 - 4\alpha c}}}{2}=\mp \frac c{|b|} = \mp \frac c{\pm b} = - \frac cb$
Okay.... so what happens to the "other" root?
For sake of argument let's assume $b > 0$. Then one of th the two roots is$\frac {-b + \sqrt{b^2 - 4ac}}{2a}$. As $a$ "gets small" $\sqrt{b^2 - 4ac}$ gets close to $b$ and this expression gets close to $0/0$ so we can use L'Hopital to get that the limit. So that was the answer we got above.
But the other root is $\frac {-b + \sqrt{b^2 - 4ac}}{2a}$. As $s$ "gets small" $\sqrt{b^2 - 4ac}$ gets close to $b$ and this expression gets close to $-2b/0$ which diverges to negative infinity.
So.. one root converges to $-\frac cb$ and the other diverges to negative infinity.
If $b=0$, then the roots are (if real), $\pm\sqrt{-c/a}$. If $c\ne0$, both roots tend to infinity (positive and negative).
If $b\ne0$, it's not restrictive to assume $b>0$ (otherwise multiply by $-1$).
If $c=0$, the roots are $0$ and $-b/a$, the latter tending to infinity ($\infty$ if $a$ approaches $0$ from the negative side, $-\infty$ if $a$ approaches $0$ from the positive side).
Assume $c\ne0$. The discriminant will be positive when $a$ belongs to a suitable (punctured) neighborhood of $0$, namely for $0<|a| Now we can rationalize the expression for the roots. First root:
$$
\frac{-b+\sqrt{b^2-4ac}}{2a}=
\frac{b^2-b^2+4ac}{2a(-b-\sqrt{b^2-4ac})}=
-\frac{2c}{\sqrt{b^2-4ac}+b}
$$
For $a\to0$ this has limit $-c/b$. This is to be expected, because the polynomial will become $bx+c$ for $a=0$. Second root:
$$
\frac{-b-\sqrt{b^2-4ac}}{2a}=
\frac{b^2-b^2+4ac}{2a(-b+\sqrt{b^2-4ac})}=
\frac{2c}{\sqrt{b^2-4ac}-b}
$$
If $a\to0$, then this has limit $\pm\infty$ (according to the sign of $c$).
I think that my last question (in the intention of a 'slightly' different issue) accidentally can be helpful. It contains alternative expressions for the roots of the quadratic equation and their behavior for different ($\mathbb{R}$) coefficients. Really universal quartic formula (Do not be afraid of the 'quartic' - there is more meaningful content about quadratic.)
I will present the reasoning for $\mathbb{R}$ coefficients, postponing their checking and possible generalization for the $\mathbb{C}$ ones.
We know that in the case of a quadratic equation $ax^2+bx+c=0$, there are two alternative formulas for (both) roots: $$x_{ \begin{array}{c} 1 \\ 2 \end{array} }=\frac{-b\mp \sqrt{\mathrm{\Delta }}}{2a}$$ and $$x_{ \begin{array}{c} 1 \\ 2 \end{array} }=\frac{2c}{-b\mp \sqrt{\mathrm{\Delta }}}$$ The second one can be obtained from the first by expanding the fraction with an complementary expression to the difference of squares. $$x_{ \begin{array}{c} 1 \\ 2 \end{array} }=\frac{-b\mp \sqrt{\mathrm{\Delta }}}{2a}\cdot \frac{-b\pm \sqrt{\mathrm{\Delta }}}{-b\pm \sqrt{\mathrm{\Delta }}}=\frac{2c}{-b\pm \sqrt{\mathrm{\Delta }}}$$ or also by using of the product Vieta formula. $$x_{ \begin{array}{c} 1 \\ 2 \end{array} }\cdot x'_{ \begin{array}{c} 2 \\ 1 \end{array} }=\frac{c}{a}$$ $$x'_{ \begin{array}{c} 2 \\ 1 \end{array} }=\frac{c}{a}\cdot \frac{2a}{-b\mp \sqrt{\mathrm{\Delta }}}=\frac{2c}{-b\mp \sqrt{\mathrm{\Delta }}}$$ (In the second method, the formulas for both roots were found in the reverse order, but it does not matter.)
The first one works well $\forall b,c$ and $\forall a\neq 0$, the second one $\forall a,b$ but $\forall c\neq 0$.
In the first case, you can also proceed by calculating the appropriate limits. Then one of the roots will go to infinity, and the other will go smoothly into the only element of the obtained linear equation. $$x_1=\frac{-b-\sqrt{\mathrm{\Delta }}}{2a}{{\mathop{\longrightarrow}\limits_{a\to 0}}}\left\{ \begin{array}{cc} \mathrm{for}\ b<0: & -\frac{c}{b} \\ \mathrm{for}\ b=0: & \infty \left[-\sqrt{-{\mathrm{sgn} c\ }\ }\right] \\ \mathrm{for}\ b>0: & -\infty \end{array} \right.$$ $$x_2=\frac{-b+\sqrt{\mathrm{\Delta }}}{2a}{{\mathop{\longrightarrow}\limits_{a\to 0}}}\left\{ \begin{array}{cc} \mathrm{for}\ b<0: & \infty \\ \mathrm{for}\ b=0: & \infty \left[\sqrt{-{\mathrm{sgn} c\ }\ }\right] \\ \mathrm{for}\ b>0: & -\frac{c}{b} \end{array} \right.$$
Wherein, only finite results make sense as solutions to the linear equation!
Which root of the quadratic equation will go into the root of the linear equation depends on the sign of the linear coefficient $b$.
Formula $x_{ \begin{array}{c} 1 \\ 2 \end{array} }=\frac{2c}{-b\mp \sqrt{\mathrm{\Delta }}}$ reproduces these results without the use of limits, but it itself requires moving to the limit when $c\ =\ 0$. $$x_1=\frac{2c}{-b+\sqrt{\mathrm{\Delta }}}{{\mathop{\longrightarrow}\limits_{c\to 0}}}\left\{ \begin{array}{cc} \mathrm{for}\ b<0: & -\frac{b}{a} \\ \mathrm{for}\ b=0: & 0 \\ \mathrm{for}\ b>0: & 0 \end{array} \right.$$ $$x_2=\frac{2c}{-b-\sqrt{\mathrm{\Delta }}}{{\mathop{\longrightarrow}\limits_{c\to 0}}}\left\{ \begin{array}{cc} \mathrm{for}\ b<0: & 0 \\ \mathrm{for}\ b=0: & 0 \\ \mathrm{for}\ b>0: & -\frac{b}{a} \end{array} \right.$$ Here all the results have meaning.
That which root of the quadratic equation will go to $0$ and which in a potentially non-zero root of the obtained equation depends on the sign of the linear coefficient $b$.
There is also the possibility of refining formulas so that the root that can be calculated directly (without the use of limits) is always transformed into the root of the obtained equation. $$x_1=\frac{-b-{\mathrm{sgn} b\ }\sqrt{\mathrm{\Delta }}}{2a}={{\left|c=0\right|}}=-\frac{b}{a}$$ $$x_2=\frac{2c}{-b-{\mathrm{sgn} b\ }\sqrt{\mathrm{\Delta }}}={{\left|a=0\right|=}}\left\{ \begin{array}{cc} \mathrm{for}\ b\neq 0: & -\frac{c}{b} \\ \mathrm{for}\ b=0: & ? \end{array} \right.$$ Of course, only the case $b\ \neq \ 0$ have meaning.
The most universal (?) set of quadratic formulas is then: $$x_1=\frac{-b-{\mathrm{sgn} b\ }\sqrt{\mathrm{\Delta }}}{2a}$$ $$x_2=\frac{2c}{-b-{\mathrm{sgn} b\ }\sqrt{\mathrm{\Delta }}}$$ and they work for all $\mathbb{R}$ values of coefficients. (I leave the generalization for $\mathbb{C}$ numbers for later.)