1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

The trinomial

$$ax^2+bx+c=0$$

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

$$ax^2+bx+c=0$$

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

Then, rewrite the equation

$$\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}=0$$

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

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

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

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

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

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

  • 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

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

Hence

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

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

$$\left(x+\frac{b}{2a}\right)^2=\frac{\Delta}{4a^2}$$

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

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

So, in all cases, the solutions are

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

Hence your equivalence holds.


Complex resolution

Again, we start with the equation

$$ax^2+bx+c=0$$

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

Then, rewrite the equation

$$\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}=0$$

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

$$\left(x+\frac{b}{2a}\right)^2-\frac{\alpha^2}{4a^2}=0$$

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

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

So the roots are

$$x=-\frac{b}{2a}\pm\frac{\alpha}{2a}$$

Or if you prefer

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

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.