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  1. #1
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    quadratic equation

    9xsquared - 6x + 1 = 0

    first use discriminate to determine whether the equation has two nonreal complex solution, one solution with a multiplicity of two, or two real solutions, and then solve the equation.

    Can someone put me in the right direction me this? thanks
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  2. #2
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    Quote Originally Posted by likewhoa View Post
    9xsquared - 6x + 1 = 0

    first use discriminate to determine whether the equation has two nonreal complex solution, one solution with a multiplicity of two, or two real solutions, and then solve the equation.

    Can someone put me in the right direction me this? thanks
    What in particular are you having trouble with? The question is very clear...


    Here you have a Quadratic Equation:

    9x^2 - 6x + 1 = 0 so a = 9, b = -6, c = 1.


    Discriminant:

    \Delta = b^2 - 4ac.

    If \Delta < 0 the solutions are complex conjugates

    If \Delta = 0 you have one repeated solution

    If \Delta > 0 you have two real solutions.


    The roots:

    x = \frac{-b \pm \sqrt{\Delta}}{2a}.
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