1. May 26, 2014

### Jhenrique

The solutions $(x_1, x_2)$ for the quadratic equation $(0=ax^2+bx+c)$:

$x_1 = \frac{-b + \sqrt{b^2-4ac}}{2a}$

$x_2 = \frac{-b - \sqrt{b^2-4ac}}{2a}$

Are true if $x$ and $a$, $b$, and $c$ $\in$ $\mathbb{C}$ ?

2. May 26, 2014

### Staff: Mentor

3. May 26, 2014

### Jhenrique

4. May 26, 2014

### symbolipoint

The formula is variable in that the formula for x depends on the numbers a, b, and c. The discriminant is b^2-4ac. The way the discriminant relates to zero tells if there are two real solutions for x, or just one real solution for x, or two complex solutions for x using imaginary numbers.

5. May 26, 2014

### Jhenrique

But the solution showed in my first post is valed for Δ=0 and Δ<0?

6. May 26, 2014

### symbolipoint

You mean delta as the discriminant? It is valid either way. If discriminant is less than zero, then , as I already said, x is not a real number. If a nonreal number makes no sense in a particular example application, then the solution for x is not valid.

7. May 26, 2014

### symbolipoint

I rechecked the last part of your first message on the topic. What I said is mostly for typical college algebra/ intermediate algebra student. The next person who responds should be a member with much more knowledge about complex numbers.

8. May 26, 2014

### Staff: Mentor

The wiki article says in plain type:

9. May 27, 2014

### Integral

Staff Emeritus
These conditions are for Real coefficients. The 2nd has no meaning if the coefficients are complex since the complex are not an ordered field. If the coefficients are complex then you would need to know how to handle the square root and do complex arithmetic, but it should yield your 2 roots.

10. May 27, 2014

### AlephZero

..which is a particular case of the http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

To answer the OP's question, just substitute the given solutions in the equation and show they are correct.

11. May 27, 2014

### Jhenrique

But $\Delta$ $\in$ $\mathbb{C}$ isn't necessary to apply the formula:

in $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ?

12. May 27, 2014

### Staff: Mentor

Thanks for the more generalized answer. The OP initial post asked if x,a,b,c could be elements of Complex numbers and so I posted the article which he did not completely agree with and so I posted the specific statement from the wiki article on the quadratic formula in response.

13. Jun 3, 2014

### Some_dude91

Solution by discriminant is not necessarily limited to real numbers, in fact it finds what values x can have,
now, which means that x can be either real or complex, the matter is where you use it. For example if you use it for real functions, it's best if you cross out a discriminant less than 0 or even complex since what you originally want is a real solution. Outside the real number limited exercises, you can use it and get the result you want. It does have some use in complex numbers, depending on what result you want to satisfy.