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Math Help - Quadratics Help!

  1. #1
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    Quadratics Help!

    I have tried the quadratic formula and also completing the square(sorta).
    Im Stuck Please help.

    (questions are not in standard form yet)

    6x2 = 4 - 5x

    4x2 + 2x = 5

    9(x - 4)2 + 1 = 0
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  2. #2
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    Re: Quadratics Help!

    Hello, Cameron!

    I have tried the Quadratic Formula and also completing the square (sorta).
    If you showed some work, we might find your errors.

    I'm stuck. Please help.

    (questions are not in standard form yet)
    Did you put them in standard form?

    [1]\;6x^2 \:=\: 4 - 5x

    We have: . 6x^2 + 5x - 4 \:=\:0

    Factor: . (2x-1)(3x+4) \:=\:0

    Therefore: . x \:=\:\tfrac{1}{2},\;\text{-}\tfrac{4}{3}




    [2]\;4x^2+ 2x \:=\:5

    We have: . 4x^2 + 2x - 5 \:=\:0

    Quadratic Formula: . x \;=\;\frac{\text{-}2 \pm \sqrt{2^2 - 4(4)(\text{-}5)}}{2(4)}

    . . . . . . x \;=\;\frac{\text{-}2\pm\sqrt{84}}{8} \;=\;\frac{\text{-}2\pm2\sqrt{21}}{8} \;=\;\frac{\text{-}1\pm\sqrt{21}}{4}




    [3]\;9(x - 4)^2 + 1 \:=\: 0

    No fancy tricks . . . just solve for x.

    We have: . 9(x-4)^2 \:=\:\text{-}1

    . . . . . . . . . (x-4)^2 \:=\:\text{-}\tfrac{1}{9}{

    n . . . . . . . . . x-4 \;=\;\pm\sqrt{\text{-}\tfrac{1}{9}}

    n . . . . . . . . . x-4 \;=\;\pm\tfrac{1}{3}i

    n . . . . . . . . . . . . x \;=\;4\pm\tfrac{1}{3}i
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