Working the quadratic formula backwards

[beetay719]

I'm really hung up on this question. I think you have to use the quadratic equation but backwards in some way.

The question is:

If 2+3i is one root of a quadratic equation with real coefficients, what is the sum of the roots of the equation?

I don't even know where to begin. Please help.

 

[Random]

Yep, you could use the quadratic backwards, just filling in what you know, but I do not think you need to.

Quadratic is [-b +AND- sqrt(b^2-4(a)(c))]/[2a]
So if 2 + 3i is one root, than 2 - 3i would be the other.
so, 2 + 3i + 2 - 3i = x
x = 4.

The equation of the parabola is y = x^2 - 4x + 13 if you would like to check. Please ask if you want to know how I got it.

Random [this is my first helping post!]

 

[beetay719]

Thanks!

Thanks a bunch!!!

 

[morson]

Yup, Random's right. If \(\displaystyle a + bi\), \(\displaystyle a,b \in\ R\) is a solution to a polynomial of degree n (\(\displaystyle \sum_{k=1}^{n+1} [A_k x^{n+1-k}]\\)), WITH REAL COEFFICIENTS, then \(\displaystyle a - bi\), its complex conjugate, is also a solution. This theorem explains why polynomials of an odd degree (for example, linears and cubics) always have at least 1 real solution.