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Math Help - Quadratic formula question

  1. #1
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    Quadratic formula question

    Good morning everyone, I am having some trobule with the following equation.
    This is what I have done so far.

    4x^2+6x-2=0
    I am using ax^2+bx+c=0
    so
    a=4
    b=6
    c=-2

    x=\frac{-(6)\pm\sqrt{6^2-4(4)(-2)}}{2(4)}
    x=\frac{-(6)\pm\sqrt{36+32}}{8}
    x=\frac{-(6)\pm\sqrt{68}}{8}
    x=\frac{-(6)\pm\sqrt{8.25}}{8}
    x=\frac{-(6)+8.25}{8},\frac{-(6)-8.25}{8}
    \frac{-14.25}{8},\frac{2.25}{8}

    Does this look right to you?
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by EricS View Post
    Good morning everyone, I am having some trobule with the following equation.
    This is what I have done so far.

    4x^2+6x-2=0
    I am using ax^2+bx+c=0
    so
    a=4
    b=6
    c=-2

    x=\frac{-(6)\pm\sqrt{6^2-4(4)(-2)}}{2(4)}
    x=\frac{-(6)\pm\sqrt{36+32}}{8}
    x=\frac{-(6)\pm\sqrt{68}}{8}
    x=\frac{-(6)\pm\sqrt{8.25}}{8}
    x=\frac{-(6)+8.25}{8},\frac{-(6)-8.25}{8}
    \frac{-14.25}{8},\frac{2.25}{8}

    Does this look right to you?
    They look right.

    Note that \sqrt{68} \approx 8.25 but if you have already rounded and converted into a decimal why not keep going?

    \frac{-14.25}{8} \approx -1.78,\frac{2.25}{8} \approx 0.28
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  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    on another note

    I perosnally prefer an exact answer that you had before...

    <br />
x=\frac{-(6)\pm\sqrt{68}}{8}<br />

    but that is just me.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    I perosnally prefer an exact answer that you had before...

    <br />
x=\frac{-(6)\pm\sqrt{68}}{8}<br />

    but that is just me.
    Or even better:
    x = \frac{-3 \pm \sqrt{17}}{4}

    Whether you want an exact answer or decimals depends, of course, on why you are solving the equation.

    -Dan
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  5. #5
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    Awsome. I cant believe I got that one right! well sort of atleast, but i guess i'm on the right track.
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