Universal quadratic formula?

Question

Is there any way to write the quadratic formula such that it works for $ac= 0$ without having to make it piecewise?

The traditional solution of $x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$ breaks when $a = 0$, and the less-traditional solution of $x = 2c / (-b \pm \sqrt{b^2 - 4ac})$ breaks when $c = 0$... so I'm wondering if there is a formula that works for both cases.

My attempt was to make the formula "symmetric" with respect to $a$ and $c$ by substituting $$x = y \sqrt{c/a}$$ to get $$y^{+1} + y^{-1} = -b/\sqrt{ac} = 2 w$$

whose solution is

$$y = -w \pm \sqrt{w^2 - 4}$$

which is clearly symmetric with respect to $a$ and $c$, but which doesn't really seem to get me anywhere if $ac = 0$.

(If this is impossible, it'd be nice if I could get some kind of theoretical explanation for it instead of a plain "this is not possible".)

Answer 1

Answering my own question, but I just realized this algorithm on Wikipedia works if we cheat a little and don't consider $\operatorname{sgn}(x) = |x| \div x$ a "piecewise" function:

$${\begin{aligned}x_{1}&={\frac {-b-\operatorname{sgn}(b)\,{\sqrt {b^{2}-4ac}}}{2a}}\\x_{2}&={\frac {2c}{-b-\operatorname{sgn}(b)\,{\sqrt {b^{2}-4ac}}}}\end{aligned}}$$

Answer 2

I think you will always have a problem, because the "missing" solution for $a=0$ is a solution at infinity. You need two answers for the quadratic, and reducing this to the linear case will give an undefined expression representing the missing root.

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Apologies, I had to get out before filling in the detail, but here is some more comment.

Taking the conventional quadratic formula with small $a$ and fixed $b,c$and examining the highest order terms, the two solutions become $-\frac {b}a$ and $-\frac cb$ plus lower order pieces.

So we have a large solution (sign depending on the sign of $a$ and which is the non-zero solution of $ax^2+bx=0$ - when $x$ is large $c$ becomes irrelevant) and a solution close to the solution of the linear equation $bx+c=0$ - where $x$ is small, the quadratic term is negligible.

Note that the second version of the solution with $c$ in the numerator has one solution which reduces to the form $\frac {2c}0$ when $a=0$, so it doesn't actually recover anything better.

To reduce a form which gives two solutions to the single solution for a linear equation requires some reduction of two solutions to one. They could become equal, but clearly that is not going to happen here (a quadratic with equal solutions is nothing like linear). Or they could disappear into a singularity or undefined space.

Answer 3

$x=(−b±\sqrt{b^2−4ac})/2a ...(1)$

Take limit $a \rightarrow 0,$ apply L'Hospital rule and you should get

$x = -c/b$ (solution of $bx + c = 0$) and another solution of infinity (neglecting since not defined).

There should not be any problem in equation (1) for the case of $c = 0$.

To find a symmetric formula for $ac$, both $a$ and $c$ must have equal significance in the quadratic equation which is not the case, that's why it's not possible.