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Solve the quadratic equation

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Solve the following equation for Z, find all solutions.

    z2 -2z + i = 0

    2. Relevant equations

    [-b(+/-) sqrt(b2-4ac)]/2a


    3. The attempt at a solution

    Using the equation above,

    z = [2 (+/-) sqrt( (-2)2 - 4 (1) (i)) ]/ 2(1)

    =[2 (+/-) sqrt ( 4 - 4i)]/2

    = [2 (+/-) sqrt ( (4) (1-i)]/2

    = [2 (+/-) sqrt(4) sqrt(1-i)]/2

    = [2 (+/-) 2 sqrt(1-i)]/2

    = 1 (+/-) sqrt (1-i)

    which means the roots are 1+ sqrt(1-i) and 1- sqrt(1-i). This is the answer that I have entered into my online assignment, but it is being marked as incorrect. Do I have an error in my calculations, or can the answer can be simplified further?
     
  2. jcsd
  3. Sep 13, 2009 #2
    Original problem:

    [tex]z^2-2z+i=0[/tex]

    Check of your solution:

    [tex]z_{1,2}=\frac{2 \pm \sqrt{4-4i}}{2}[/tex]

    [tex]z_{1,2}=\frac{2 \pm 2\sqrt{1-i}}{2}[/tex]

    [tex]z_{1,2}=1 \pm \sqrt{1-i}[/tex]

    Another approach:

    [tex](z-1)^2-1+i=0[/tex]

    [tex](z-1)^2=1-i[/tex]

    [tex]z-1=\pm \sqrt{1-i}[/tex]

    [tex]z = 1 \pm \sqrt{1-i}[/tex]
     
  4. Sep 13, 2009 #3
    So I have it right then. I guess the site is being picky with the answer format or something. Thank you!
     
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