Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help solving an equation

  1. Jun 25, 2004 #1
    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. jcsd
  3. Jun 25, 2004 #2
  4. Jun 25, 2004 #3
    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?
  5. Jun 25, 2004 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Proof lies in completing the square.

    [tex]ax^2 + bx + c = 0 = {x^2} + \frac {b} {a} x + \frac {c} {a} [/tex]
    [tex] (x + \frac {b} {2a})^2 - \frac {b^2} {4a^2} +\frac {c} {a} =0 [/tex]

    Take constant terms to other side, find the square root and subtract b/2a to get the quadratic formula.
  6. Jun 25, 2004 #5
    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!
  7. Jun 25, 2004 #6
    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!
  8. Jun 25, 2004 #7


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to PF kuenmao...check out the LaTex post under General Physics.
  9. Jun 25, 2004 #8
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook