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Math Help - Need help with quadratic equation.

  1. #1
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    Need help with quadratic equation.

    Can anyone show me how to solve this using the quadratic equation?

    e^3=x^2 + 6x + 8

    It says in the solution you must group e^3 with 8. Therefore making the problem...

    x^2 + 6x -(e^3-8)=0

    I'm just stuck on how you put this into the quadratic equation.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Need help with quadratic equation.

    Notice e^3 is also a constant so rewrite the equation as:
    x^2+6x+(8-e^3)=0

    Now use the quadratic formula.

    EDIT:
    I notice you have modified your post.
    Calculate D=b^2-4ac=6^2-4.1.(8-e^3)=...

    Afterwards you just have to substitute the discriminant into:
    x_1,x_2=\frac{-b\pm \sqrt{D}}{2a}

    Where x_1 and x_2 are the two solutions of the equation.
    Last edited by Siron; August 31st 2011 at 08:14 AM.
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  3. #3
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    Re: Need help with quadratic equation.

    Can you possibly show the work to solve for x? The answer key says....

    x= -3 + sqrt(1+e^3)

    When I work out the problem, I get....

    x= -3 + sqrt(4e^3)

    What am I doing incorrectly?
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Need help with quadratic equation.

    If you calculate the discriminant you get:
    D=(6)^2-4\cdot (8-e^3)=36-32+4e^3=4+4e^3=4(1+e^3)
    So the solutions are:
    x_1=\frac{-6+\sqrt{4(1+e^3)}}{2}=\frac{-6}{2}+\frac{2\sqrt{1+e^3}}{2}=-3+\sqrt{1+e^3}
    x_2=\frac{-6-\sqrt{4(1+e^3)}}{2}=-3-\sqrt{1+e^3}
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  5. #5
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    Re: Need help with quadratic equation.

    Quote Originally Posted by bseymore View Post
    Can anyone show me how to solve this using the quadratic equation?

    e^3=x^2 + 6x + 8

    It says in the solution you must group e^3 with 8. Therefore making the problem...

    x^2 + 6x -(e^3-8)=0

    I'm just stuck on how you put this into the quadratic equation.
    \displaystyle \begin{align*} x^2 + 6x + 8 &= e^3 \\ x^2 + 6x &= e^3 - 8 \\ x^2 + 6x + 3^2 &= e^3 - 8 + 3^2 \\ (x + 3)^2 &= e^3 + 1 \\ x + 3 &= \pm \sqrt{e^3 + 1} \\ x &= -3 \pm \sqrt{e^3 + 1} \end{align*}
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