Larger Theory for root formula

Question

Consider the quadratic equation:

$$ax^2 + bx + c = 0$$

and the linear equation:

$$bx + c = 0$$.

We note the solution of the linear equation is

$$x = -\frac{c}{b}.$$

We note the solution of the quadratic equation is

$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$.

Suppose we take the limit as $a$ approaches 0 on the quadratic equation: ideally we should get the expression $-\frac{c}{b}$ left over, but this is clearly not the case:

This means that the quadratic formula does not generalize the linear formula, it is instead streamlined to only solve quadratic equations.

What would the general formula be? Basically what is the formula such that if $a$ is non-zero is equivalent to the quadratic equation and if $a = 0$ breaks down to the linear case?

Answer 1

Hint $\ $ First rationalize the numerator, then take the limit as $\rm\:a\to 0.$

$$\rm \frac{-b + \sqrt{b^2 - 4ac}}{2a}\ =\ \frac{2c}{-b -\sqrt{b^2 - 4ac}}\ \to\ -\frac{c}b\ \ \ as\ \ \ a\to 0 $$

Remark $\ $ The quadratic equation for $\rm\,\ z = 1/x\,\ $ is $\rm\,\ c\ z^2+ b\ z + a = 0\,\ $ hence

$$\rm z\ =\ \dfrac{1}{x}\ =\ \dfrac{-b \pm \sqrt{b^2-4\:a\:c}}{2\:c} $$

Inverting the above now yields the sought limit as $\rm\:a\to 0,\:$ so effectively removing the apparent singularity at $\rm\: a = 0\:.\ $

Answer 2

Yes you do. But you can't just take the limit as $a\to 0$, because that gives an indeterminate form $0/0$. You need to do some algebra first:

$$\lim_{a\to 0}\frac{-b+\sqrt{b^2-4ac}}{2a}$$

I use $+$ because assuming $b>0$ (so $\sqrt{b^2}=b$) simplifies the problem. In the case where $b$ is negative, the other root will be the one to converge, and the case can be handled in the same way. Considering this as an equation in $a$, we use L'Hopital's rule (probably overkill, but works):

$$\lim_{a\to 0}\frac{\frac{-2c}{\sqrt{b^2-4ac}}}{2}=-c/|b|=-c/b$$

If $b>0$, this gives $-c/b$. If $b<0$, choosing the other root would remove the negative in the numerator, while re-introducing it in $|b|$. So the formula is $-c/b$ in either case.

Note that you can expect this looking at this geometrically. If you fix $b$ and $c$ and let $a$ vary, $ax^2+bx+c\to bx+c$. The parabola will appear to steadily flatten out to approach a line. One root will converge (to the root of the line), while the other root will zoom off to $\pm\infty$. And this is the behaviour we see in the equation as well, once we tease things out with a little calculus.