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Math Help - quadratic formula

  1. #1
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    quadratic formula

    3x^2 - 4x + 1 =0
    This is what I did.
    -(-4) = -4^2 - 4(3)(1) / 2(3)
    4 = 16 - 12 / 6
    4 = 4/6 = 1/3
    Another answer is 1. How do they get 1.

    x(x+6) = -11 - x
    x^2 + 6x + x = -11
    -6 = 6^2 - (4)(1)(1) / 2(1)
    6 = 36 - 4 / 2
    6 = 32 / 2
    What do I do next?
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  2. #2
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    I'm not sure why there are extra equal signs ....

    3x^{2} -4 x + 1 = 0

    x = \frac{-(-4) \: {\color{red} \pm}\: \sqrt{(-4)^{2} - 4(3)(1)}}{2(3)}

    x = \frac{4 \:{\color{red} \pm}\: \sqrt{16 - 12}}{6}

    x = \frac{4 \:{\color{red} \pm}\: \sqrt{4}}{6} = \frac{4 \pm 2}{6}

    x = \frac{4+2}{6} = 1 \quad \text{and} \quad x = \frac{4 - 2}{6} = \frac{1}{3}
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  3. #3
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    How do you do the second problem?
    Last edited by rowdy3; April 24th 2008 at 04:42 PM. Reason: spelling error
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  4. #4
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    x(x+6) = -11 - x
    x^{2} + 6x = -11 - x
    x^{2} + 7x + 11 = 0

    Apply the quadratic formula again.
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  5. #5
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    Thanks I got it.
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