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Math Help - A question. Finding the intercepts in -5x^2+20x+2... Having some issues.

  1. #1
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    A question. Finding the intercepts in -5x^2+20x+2... Having some issues.

    So I've got -5x^2+20x+2, and have to determine the x intercepts.

    I can use partial factoring to get -5x(x-4)+2

    from which I can get

    0=-5x
    0/-5=x
    0=x
    AND

    0=x-4
    4=x

    Which gives me 0 and 4.

    My issues that when I enter -5x^2+20x+2 into a graphing calculator, the x-intercept coordinates I am given are -0.098 and 4.098. I'd like to know why. Have I done something wrong?

    Thanks so much for the help!
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: A question. Finding the intercepts in -5x^2+20x+2... Having some issues.

    The x-interceopt are the values for x where -5x^2+20x+2 = 0. The mistake you made is going from -5x(x-4)+2=0 to -5x(x-4) = 0; what happened to the 2? You should always double check your answers to see if they work. If you plug x=0 or x = 4 into the original equation you get:

    -5(0^2)+20(0)+2 = 0,
     2 = 0

    and:

    -5(4^2)+20(4)+2 =0,
    2 = 0

    so clearly these answers are not right.

    Easiest way to find the roots is to apply the quadratic equation:

    x = \frac {-20 \pm \sqrt {20^2-4(-5)(2)}} {2(-5)}
    Last edited by ebaines; August 21st 2012 at 12:37 PM.
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  3. #3
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    Re: A question. Finding the intercepts in -5x^2+20x+2... Having some issues.

    Quote Originally Posted by OverlyFrustrated View Post
    So I've got -5x^2+20x+2, and have to determine the x intercepts.
    Well the roots are 2\pm {\sqrt{110}/{5}
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    Re: A question. Finding the intercepts in -5x^2+20x+2... Having some issues.

    Hi there,

    Thanks for taking the time to respond.

    I've read the question I was basing this answer off of again and realize that there was no C value in the example I was provided.

    In addition, it appears I was doing all kinds of bad math with the quadratic formula.

    Thank you very much for your help.
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