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Math Help - another quadratic forumla question

  1. #1
    Junior Member
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    another quadratic forumla question

    I have the equation -2x^2+3x+7=0

    So using the quadratic forumla I have

    -3 plus or minus sqrt(3^2-4(-2)(7))/-4

    = -3 plus or minus sqrt(65)/-4

    Now, in the solutions manual it has the solutions as

    (3 - sqrt(65))/4 , (3+sqrt(65))/4

    Just wondering, but why did the denominator of the solutions become positive? and why did the the 3 also become positive? Thanks!!!

    Thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathfailure View Post
    I have the equation -2x^2+3x+7=0

    So using the quadratic forumla I have

    -3 plus or minus sqrt(3^2-4(-2)(7))/-4

    = -3 plus or minus sqrt(65)/-4

    Now, in the solutions manual it has the solutions as

    (3 - sqrt(65))/4 , (3+sqrt(65))/4

    Just wondering, but why did the denominator of the solutions become positive? and why did the the 3 also become positive? Thanks!!!

    Thanks
    the solutions manual multiplied the answers by -1/-1 (which is 1 so it doesn't change anything).

    alternately, the solutions manual could multiply the original equation through by -1 to get the coefficient of x^2 positive (that is, the equation becomes 2x^2-3x-7=0). doing that, you would get the solution the manual has right off the bat
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