# Help solving an equation

1. Jun 25, 2004

### Ed Quanta

Can someone help me solve this equation which for some reason is giving me trouble?

z^2+z+1=0 where z is a complex number, or if it makes it easier we can write

(x,y)(x,y) + (x,y) + (1,0)= (0,0)

2. Jun 25, 2004

### fourier jr

3. Jun 25, 2004

### Ed Quanta

I was trying to do it by separating equations. Anyway, thank you, do you know where I can find an online proof of the quadratic formula?

4. Jun 25, 2004

### Gokul43201

Staff Emeritus
Proof lies in completing the square.

$$ax^2 + bx + c = 0 = {x^2} + \frac {b} {a} x + \frac {c} {a}$$
$$(x + \frac {b} {2a})^2 - \frac {b^2} {4a^2} +\frac {c} {a} =0$$

Take constant terms to other side, find the square root and subtract b/2a to get the quadratic formula.

5. Jun 25, 2004

### kuenmao

I'll just type that for you: (the "^" means "to the power")

a x^2 + b x + c = 0
x^2 + (b x)/a + c/a = 0 (divide both sides by a, since a is not zero)
x^2 + (b x)/a = -c/a
x^2 + (b x)/a + (b/2a)^2 = -c/a + (b^2)/(4 a^2) (Add (b/2a)^2 to both sides)
By the identity a^2 + 2ab + b^2 = (a+b)^2, we have
(x + b/2a)^2 = -c/a +(b^2)/(4 a^2)
(x + b/2a)^2 = (b^2-4ac)/(4 a^2)
x + b/2a = sqrt.[b^2-4ac] / 2a or -sqrt.[b^2-4ac] / 2a
x = (-b + sqrt.[b^2-4ac]) / 2a or (-b - sqrt.[b^2-4ac]) / 2a

And so you have the quadratic formula. Hope that helps!

6. Jun 25, 2004

### kuenmao

whoops, I am new here and just realized that you could create the equations yet...sorry about making that long chunk in the previous post!

7. Jun 25, 2004

### Gokul43201

Staff Emeritus
Welcome to PF kuenmao...check out the LaTex post under General Physics.

8. Jun 25, 2004

### robert Ihnot

the roots of unity

Another way to look at that is (z-1)(z^2+z+1) = z^3 -1. Thus we are talking about the three roots of 1. Those roots are cos(k(120)) + isin(k(120)), for k=1,2,3. This actually is a better way to do it from the standpoint of insight into the roots of unity. However it is not a general method for solving the quadratic.

Last edited: Jun 25, 2004