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Math Help - using the quadratic formula

  1. #1
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    using the quadratic formula

    Use the quadratic formula to solve the equation

    x^2 - x = -7

    x^2 - x + 7 = -7 + 7

    x^2 - x + 7 = 0

    a= 1, b= -1, c= 7

    - (-1) + or - sqrt (-1)^2-4(1)(7)/2(1)

    1 + or - sqrt 1-28/ 2
    1 + or - sqrt -27 /2

    am I on the right track here?
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  2. #2
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    Quote Originally Posted by Just4 View Post
    Use the quadratic formula to solve the equation

    x^2 - x = -7

    x^2 - x + 7 = -7 + 7

    x^2 - x + 7 = 0

    a= 1, b= -1, c= 7

    - (-1) + or - sqrt (-1)^2-4(1)(7)/2(1)

    1 + or - sqrt 1-28/ 2
    1 + or - sqrt -27 /2

    am I on the right track here?

    You are correct . You can continue from here to find its complex roots .
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  3. #3
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    I thought this was the finally answer

    1 + or - sqrt -27/2

    could you explain to me how I would find the complex roots?
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  4. #4
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    Quote Originally Posted by Brama Bull View Post
    I thought this was the finally answer

    1 + or - sqrt -27/2

    could you explain to me how I would find the complex roots?
    First, be careful with your parentheses! Most people would read "1 + or - sqrt -27/2" as 1\pm\sqrt{\frac{-27}{2}} but what you mean is \frac{1\pm\sqrt{27}}{2} which, without LaTex, should be written as (1 + or - sqrt(27))/2.

    Now, for the square root: \sqrt{-27}= \sqrt{(-1)(27)}= \sqrt{-1}\sqrt{27}. The square root of -1 is, by definition, "i". So \sqrt{-27}= i\sqrt{27}. It is also true that 27= 9(3) so \sqrt{27}= \sqrt{9}\sqrt{3}= 3\sqrt{3}. Your teacher might insist that you write the answer as \frac{1\pm 3i\sqrt{3}}{2} and, in any case, it is better to do so.
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