I have been upto proving the following

$$(\forall x\in \mathbb C, ax^2 + b x + c = 0) \land(a\neq0)\Leftrightarrow (x = {\frac {-b \pm \sqrt{b^{2}-4ac}} {2a}})$$

Due to equality we need to proof bi implication,i.e. $a\Rightarrow b$ and $a\Leftarrow b$.It involves just simple arithmetic to go from roots to quadratic equation, but the other way is bit tricky. This is what I have done so far
$$(\forall x\in \mathbb C, ax^2 + b x + c = 0) \Rightarrow (x = {\frac {-b \pm \sqrt{b^{2}-4ac}} {2a}})$$

$$(\forall x\in \mathbb C, c = -ax^2 - b x ) \Rightarrow (2ax = {-b \pm \sqrt{b^{2}-4ac}} )$$

$$(\forall x\in \mathbb C, c = -ax^2 - b x ) \Rightarrow (2ax + b = {\pm \sqrt{b^{2}-4ac}} )$$

Using value of $c$ we get complete square, equation looks like this

$$(\forall x\in \mathbb C, c = -ax^2 - b x ) \Rightarrow ((2ax + b) = {\pm \sqrt{(2ax+b)^2} })$$

On this forum I have found discussion on forms like $$(x = {\pm \sqrt{x^2} })$$ For complex numbers, I had an impression that it is not true. But it raises question on the goal I am trying to proof.So, if I summarize, I need comments regarding

1-is goal legal?

2-If legal then how to deal with resulting form?

3- If not (1), then looking for some alternative approach to handle it.

thanks in advance

  • $\begingroup$ Do you have the fundamental theorem of algebra at your disposal? This would settle equivalence once you've shown that the formula gives the two roots. (Watch out for $b^2-4ac = 0$, though). $\endgroup$ – AlexR Sep 12 '14 at 8:44
  • $\begingroup$ At some point you want to assume the coefficient of $x^2$ is not zero. $\endgroup$ – Gerry Myerson Sep 12 '14 at 9:19
  • $\begingroup$ @GerryMyerson Right I have edited the question thanks for mentioning $\endgroup$ – Asad Sep 12 '14 at 10:39

The trinomial


May be solved when $a,b,c$ are complex numbers, and it works fine with real $a,b,c$, since $\Bbb R\subset\Bbb C$.

However, we'll start with the real case.

Real resolution

We start with the equation


Where $a\neq0$, and $a,b,c\in\Bbb C$.

Then, rewrite the equation


Then, according to the sign of $\Delta=b^2-4ac$, there are three possibilities:

  • If $\Delta=0$, the equation amounts to


It has thus one double root, $x=-\frac{b}{2a}$.

But since $\Delta=0$, we can also write


  • If $\Delta>0$, then $\frac{b^2-4ac}{4a^2}$ is positive, so it's the square of number $\frac{\sqrt{\Delta}}{2a}$, and we can write

$$\left(x+\frac{b}{2a}\right)^2-\left(\frac{\sqrt{\Delta}}{2a}\right)^2=0$$ $$\left(x+\frac{b}{2a}+\frac{\sqrt{\Delta}}{2a}\right)\left(x+\frac{b}{2a}-\frac{\sqrt{\Delta}}{2a}\right)=0$$



  • If $\Delta<0$, then the equation has no real solution, since it would amount to


Where the left-hand side is $\geq0$, whereas the right-hand side is $<0$.

However, we can still write that $\Delta$ is the square of complex number $i\sqrt{4ac-b^2}$, hence we can write

$$\left(x+\frac{b}{2a}\right)^2-\left(\frac{i\sqrt{-\Delta}}{2a}\right)^2=0$$ $$\left(x+\frac{b}{2a}+\frac{i\sqrt{-\Delta}}{2a}\right)\left(x+\frac{b}{2a}-\frac{i\sqrt{-\Delta}}{2a}\right)=0$$

$$x=\frac{-b\pm i\sqrt{-\Delta}}{2a}$$

But, with the convention that $\sqrt{t}=i\sqrt{-t}$ when $t<0$, we can again write


So, in all cases, the solutions are


Hence your equivalence holds.

Complex resolution

Again, we start with the equation


Where $a\neq0$, and $a,b,c\in\Bbb C$.

Then, rewrite the equation


Now, a complex number $z$ has always a square root (two distinct ones if $z\neq0$, see for example here). We go on with a square root $\alpha$ of $b^2-4ac$, that is, $\alpha^2=b^2-4ac$, so that the equation can be written


It's a difference of squares, and we know that $A^2-B^2=(A+B)(A-B)$, so


So the roots are


Or if you prefer


Important note: there is no canonical way to write that $\sqrt{t}$ is a specific square root of $t\in\Bbb C$, hence this symbol is meaningless. However, since it appears with a $\pm$, we will always work with both roots, so using the symbol here is harmless. Nevertheless, it should be avoided.

  • $\begingroup$ S@Jean-Claude Arbaut Thanks for time, I did try this way but I have little problem. As it is allowed to replace $$(\sqrt{{b^2-4*a*c\over 4*a^2}})^2={b^2-4*a*c\over4*a^2}$$ But on removing square root it will give out $2*abs(a)$ which I doubt If I can handle,i.e to replace with "$a$" again. $\endgroup$ – Asad Sep 12 '14 at 10:36
  • $\begingroup$ @Asad No, $\sqrt{x^2}=|x|$, for real $x$, and $(\sqrt{x})^2=x$, for real $x\geq0$. More generally, if $\alpha$ is a square root or a (real or complex) number $\beta$, then you always have $\alpha^2=\beta$. That's the definition of a square root! $\endgroup$ – Jean-Claude Arbaut Sep 12 '14 at 11:34
  • $\begingroup$ @Asad Yes, sorry, I didn't pay enough attention. You can use the usual resolution for the real case then. But actually it's simpler to say $a,b,c$ are also complex numbers ($\Bbb R\subset\Bbb C$), and do the complex case of the resolution, since you don't have to manage different cases for $b^2-4ac>0$ and $b^2-4ac<0$ and $b^2-4ac=0$. I'll add this to the answer, though. $\endgroup$ – Jean-Claude Arbaut Sep 12 '14 at 12:10

The cleanest way to show that is probably to set $$z_{\pm} := \frac{-b\pm (b^2-4ac)^{\frac12}}{2a}$$ Where $a,b,c\in\mathbb C$ and see that $$(z-z_+)(z-z_-) = az^2 + bz + c$$ And finally use the FTA to exclude the existence of any further roots.

  • $\begingroup$ Thanks for hint, please refer some material, it would be appreciated if give some book(s)/publication(s) names to further pursue the idea. $\endgroup$ – Asad Sep 12 '14 at 9:41
  • $\begingroup$ @Asad What do you have in mind? This is about the FTA. $\endgroup$ – AlexR Sep 12 '14 at 11:27

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