# If $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ show that $x=-c/b$ when $a=0$

OK, this one has me stumped. Given that the solution for $ax^2+bx+c =0$ $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\qquad(*)$$

How would you show using $(*)$ that $x=-c/b$ when $a=0$

(Please dont use $a=0$ hence, $bx+c=0$.)

• Is this a real question or just some puzzle? Commented Aug 30, 2011 at 19:46
• @FoolForMath Actually, I don't know for sure. This was stated as a puzzle in a math department whiteboard on the hallway Commented Aug 30, 2011 at 19:47
• The quadratic formula is used to find the solutions of a quadratic equation if $a=0$ then $ax^2+bx+c =0$ is not quadratic anymore it is rather a linear equation,so unless it is possible to quadratic formula for linear equations the above question seems doesn't makes much sense to me. Commented Aug 30, 2011 at 19:51
• You can often find this discussed in older algebra texts (roughly, before the 1920s). Look up "infinite roots" in the index of any such books, or look at the following google book search results: google.com/… Commented Aug 30, 2011 at 20:11
• Kind of a weird question in a Mathematics department. Roots of polynomials can display an extreme sensitivity to small perturbations of the coefficients. Commented Aug 30, 2011 at 20:18

You don't. The quadratic formula only works if $a\neq 0$ (remember, if you try dividing by $0$, the universe explodes).

If $a=0$, then the equation is $$0x^2 + bx + c = 0$$ or equivalently, $$bx + c = 0.$$ So, solve that equation for $x$. If $b=0$, then it's always false if $c\neq 0$, and always true if $c=0$.

And if $b\neq 0$, then...

Given your comment... you could look at limiting behavior. Then \begin{align*} \lim_{a\to 0}\frac{-b\pm\sqrt{b^2-4ac}}{2a} &= \lim_{a\to 0}\frac{b^2-(b^2-4ac)}{2a\left(-b\mp\sqrt{b^2-4ac}\right)}\\ &= \lim_{a\to 0}\frac{2c}{-b\mp\sqrt{b^2-4ac}}. \end{align*}

As $a\to 0$, $\sqrt{b^2-4ac}\to |b|$. If $b\gt 0$, only the minus sign in the denominator makes sense, giving you $\frac{2c}{-2b} = -\frac{c}{b}$. If $b\lt 0$, only the plus sign in the denominator makes sense, giving you again $\frac{2c}{-2b} = -\frac{c}{b}$. (The "other root" goes to $\infty$ or $-\infty$, and so is "out of sight, out of mind..."; unless $c=0$; but if $c=0$, then the original function has one value equal to $0$ and the other to $-\frac{b}{a}$; the latter goes to $\pm\infty$ as $a\to 0$, and the former has limit $0=-\frac{c}{b}$).

And if $b=0$, then of course the whole thing doesn't makes sense unless $c=0$, in which case $x$ can be anything.

• This was stated as a puzzle in a math department whiteboard on the hallway (that I visited today).I don't know if this is what they inteded... Commented Aug 30, 2011 at 19:39
• @kuch: It sounds to me as if this is what was intended. :) Commented Aug 30, 2011 at 19:56
• Re: For the negative sign, the limit is the same. If I take the root with the negative sign in the case $b > 0$, the numerator $-b - \sqrt{b^2-4ac} \to -2b$, which is bounded away from $0$ as $a \to 0$. So the limit seems not to exist. Commented Aug 30, 2011 at 19:57
• @Srivatsan: Rewrote and did something else instead. Indeed, it was a little too handwavy before. Commented Aug 30, 2011 at 20:01
• @Billy I was referring to the first version of the answer. Commented Aug 30, 2011 at 20:03

Hint  The quadratic equation for $$\rm\, z = 1/x\,$$ is $$\rm\ c\, z^2+ b\, z + a = 0\$$ [reverse/reciprocal poly], so

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

Inverting the above now yields the sought limits as $$\rm\:a\to 0.\,$$ Note that this yields the same result as obtained by Arturo by rationalizing the numerator. Applied to the quadratic formula it removes the apparent singularity at $$\rm\: a = 0.\,$$

Remark  Coincidentally, I mentioned this method yesterday in a comment to an answer by DJC. My linked comment was posted after rote application of this method to a handful of problems posed by Jordan Carlyon on calculating limits of radical expressions without using any knowledge of derivatives. Thus this question is solvable using insight gained from standard (pre-derivative) limit exercises frequently posed in calculus courses. However, most students don't notice the application of such methods in this context (that seems to be the case considering that there are many questions like this on other math forums at this level, e.g. sci.math).

Note  As Dave Renfro mentioned in a comment, the inversion technique is mentioned in many older textbooks, e.g. Chrystal's famous Algebra. In older times this idea was more natural since there was more widespread knowledge of projective geometry - which lies at the heart here.

• See here for more. Commented May 13 at 19:32

If the question is "How would you show using (*) that $x=−c/b$ when $a=0$?" I'm inclined to say "I wouldn't (since you can't divide by $0$)." If $a=0$, then $ax^2+bx+c=0$ becomes $bx+c=0$, and it's pretty easy to show using algebra that in that case $x = -c/b$, provided that $b\neq0$.

But one could speak of $$\lim_{a\to 0} \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$$ in other words, try to show that $\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$ approaches $-c/b$ as $a$ approaches $0$. I would do the $+$ case and the $-$ case separately, and begin by rationalizing the numerator. You should get a "$+b^2$" term and a "$-b^2$" term canceling each other out. Finally, the numerator and denominator, when simplified and then factored, should both be divisible by $a$.

It could also be done by using a rationalizing substitution.

• that is what I tried. I tried taking limits and making power series approximations with $b^2>4ac$ Commented Aug 30, 2011 at 19:53

One of the two solutions approaches $-\frac{c}{b}$ in the limit as $a\rightarrow 0$. Assuming $b$ is positive, apply L'Hospital's rule to $$x=\frac{-b+\sqrt{b^2-4ac}}{2a}$$ If $b$ is negative, work with the other solution.

(And as $a\rightarrow 0$, the second solution approaches $\pm\infty$.)

• Note that to compute the limit as $\:a\to 0\:$ does not require using L'Hospital's rule, or interpreting the limit as a derivative. Rather, it suffices to apply simple pre-derivative tricks for removing the apparent singularity, such as rationalizing the numerator. See my answer here for more on this. Commented Aug 30, 2011 at 21:46