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x^2-4x+32=0.
would the final answer be 2 +or- 2 sqrt7 ?

  • math -

    x^2-4x+32=0

    a =1
    b= -4
    c= 32

    4 +or- sq root (-4)^2 -4(1)(32)
    4 +or- sq root 16-128
    4 +or- sq root -112

    you can't take the sq root of a (-) number so you can't solve this :/

    How did you get 2 +or- 2 sqrt7.
    Perhaps I went wrong somewhere.

  • math -

    hmmm. what's b^2-4ac? 16-128 = -112

    So, x = 2 ± 2√(-7) = 2 ± 2√7 i

    gotta watch those pesky minus signs

  • math -

    the original question was
    x^2-4x=32

    can you help me solve?

  • math -

    as I suspected, that makes it x^2-4x-32=0
    since 32=8*4,

    (x-8)(x+4) is the factorization

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