$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ 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, $bx+c=0$.)
 A: Hint $\ $ 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 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 $\rm\: a = 0.\,$ 
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.
A: 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.
A: 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.
A: 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 $\pm\infty$.)
