Quadratic Formula Solutions for Complex Numbers

[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 $c$ $\in$ $\mathbb{C}$ ?

 

[jedishrfu]

yes:

http://en.wikipedia.org/wiki/Quadratic_equation

 

[Jhenrique]

yes:

http://en.wikipedia.org/wiki/Quadratic_equation

In this article, is sayd that the formula is variable and is dependent of the discriminant...
http://en.wikipedia.org/wiki/Quadratic_equation#Discriminant

 

[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.

 

[Jhenrique]

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.

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

 

[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.

 

[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.

 

[jedishrfu]

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.

The wiki article says in plain type:

A quadratic equation with real or complex coefficients has two solutions, called roots. These two solutions may or may not be distinct, and they may or may not be real.

Doesn't that answer your question?

 

[Integral]

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

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.

 

[AlephZero]

The wiki article says in plain type:

A quadratic equation with real or complex coefficients has two solutions, called roots. These two solutions may or may not be distinct, and they may or may not be real.

..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.

 

[Jhenrique]

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

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

 

[jedishrfu]

..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.

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.

 

[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.