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Math Help - Quad. Eq. Need a check+help

  1. #1
    Raj
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    Quad. Eq. Need a check+help

    Determine the roots of the following. Simplified radical form.

    A) \frac{x-1}{x+4} - \frac{x}{x-3}=9

     (x-1)(x-3) - (x)(x+4) = 9(x+4)(x-3)
     x^2-4x+3 - x^2+4x = 9x^2+9x-108
     -8x+3 = 9x^2=9x^2+9x-188
     9x^2+17x-111=0

    Then i used the quad formula. My only question is, in the above second line do i distribute the - to the +4x? Or do i keep it positive? I went and made it -4x to get the below answer.

    x = \frac{-17 \pm \sqrt{4285}}{18}<br />

    and then if its not any trouble

    7. The quadratic function y=9x^2+bx+c has no x-intercepts but the zeros of the function are \frac{7 \pm \sqrt{-6}}{3}. Determine the values of b and c. Show all work.

    I don't know whats going on here so explain if you feel like it
    Last edited by Raj; October 15th 2007 at 06:54 AM.
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  2. #2
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    Hello, Raj!

    Determine the roots of the following.

    A)\;\;\frac{x-1}{x+4} - \frac{x}{x-3}=9

     (x-1)(x-3) - x(x+4) \:= \:9(x+4)(x-3)

     x^2-4x+3 - x^2+4x \:= \:9x^2+9x-108

     9x^2+17x-111=0

    Then i used the quad formula: . x = \frac{-17 \pm \sqrt{4285}}{18}
    You did great!

    7. The quadratic function y\:=\:9x^2+bx+c has zeros: . \frac{7 \pm \sqrt{\text{-}6}}{3}.
    Determine the values of b and c.
    We can run the problem "backwards" . . .

    The roots are: . \begin{array}{ccc}x & = & \frac{7 + \sqrt{\text{-}6}}{3} \\ x & = & \frac{7 - \sqrt{\text{-}6}}{3}\end{array}

    This gives us: . \begin{array}{ccccccc}x - \frac{7 + \sqrt{\text{-}6}}{3} \: = \: 0 & \Rightarrow & 3x - 7 + \sqrt{\text{-}6} \: = \: 0 \\ x - \frac{7-\sqrt{\text{-}6}}{3} \:= \:0 & \Rightarrow & 3x - 7 - \sqrt{\text{-}6} \:= \:0\end{array}

    Multiply the two equations: . \left[(3x-7) + \sqrt{\text{-}6}\right]\left[(3x-7) - \sqrt{\text{-}6}\right] \:=\:0

    . . and we have: . (3x-7)^2 - (\sqrt{\text{-}6})^2 \:=\:0\quad\Rightarrow\quad 9x^2 - 42x + 49 - (\text{-}6) \:=\:0


    Hence, the function is: . y \;=\;9x^2 - 42x + 55

    . . Therefore: . \boxed{b \,= \,-42,\;c \,= \,55}

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