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?
On
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.
On
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.
On
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 $0<|a|<b^2/|4c|$, so we have two roots.
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 $c$).
On
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.
On
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.
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 $\pm \infty$.