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

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 $\pm\infty$.)
