Through a simple mathematical substitution, I have stumbled upon an alternative formula for solving a quadratic equation:

$$x=\frac{2c}{-b \pm \sqrt{b^{2}-4ac}}$$

(Please refer to my formula derivation here; this is my first question, so cannot include images yet).

I am not aware if this has been mentioned elsewhere.

I was just curious for a formula that reduces to the linear solution ($x=-\frac{c}{b}$) for the standard quadratic $ax^2 + bx + c = 0$, if we allow the case $a=0$. The standard solution formula $x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$crashes for $a=0$ because division by zero is not allowed.

This alternative formula $x=\frac{2c}{-b \pm \sqrt{b^{2}-4ac}}$covers the case $a=0$, wherein it reduces to the linear equation $(bx + c = 0)$ solution ($x=-\frac{c}{b}$) for the negative square root of the discriminant, but the positive square root of the discriminant needs to be ignored/is a problem in that case.

What are your thoughts? Can we arrive at a general formula for a quadratic equation which includes all the cases ($a=0$ and/or $b=0$ and/or $c=0$)?

  • 1
    $\begingroup$ It’s a very interesting thought process, at least. $\endgroup$
    – littleO
    Mar 21 at 7:35
  • 2
    $\begingroup$ These are the same formula, arrived at by multiplication by conjugate $\endgroup$ Mar 21 at 7:37
  • 2
    $\begingroup$ One comment is that if $c=0$, I think the alternative formula could miss a nonzero solution. $\endgroup$
    – littleO
    Mar 21 at 7:40
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    $\begingroup$ The standard solution formula does not really crash when $a = 0$. Assume that $b > 0$. On the one hand $|x_-| \to + \infty$ as $a \to 0$ (but this is expected, since there is only one solution). On the other hand $x_+ \to -c/b$ as $a \to 0$ (as can be checked by writing $\sqrt{b^2-4ac} \approx b(1 - 2 ac / b^2)$), so the formula can be extended by continuity. $\endgroup$
    – cs89
    Mar 21 at 8:51
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    $\begingroup$ FYI, this alternate form is often called the Citardauq Formula (not sure when this name began -- possibly this 1976 paper), and one can find it in algebra texts (for numerically calculating roots when the quadratic coefficient is close to zero) at least as far back as the mid 1800s. With the rise of electronic computers (since mid 1940s), it's fairly widely known in CS. $\endgroup$ Mar 21 at 9:13

1 Answer 1


Look at the homogeneous case where we are working with points $(x:y)$ on the complex projective line and consider the homogeneous equation $$ax^2+bxy+cy^2=0.$$ The solutions are $$(x:y)=(-b \pm\sqrt{b^2-4ac}:2a)$$ and this works as long as $a,b,c$ are not all $0.$

  • 1
    $\begingroup$ [+1] This is more or less the algebraic equivalent of @cs89's comment invoking limits. $\endgroup$
    – ronno
    Mar 21 at 11:27
  • 1
    $\begingroup$ Hardly understood anything from this answer. Kindly forgive my mathematical incompetency. $\endgroup$ Mar 21 at 18:06

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