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Math Help - Find the Positive Root Of The Following Equation

  1. #1
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    Find the Positive Root Of The Following Equation

    Positive root of the following equation:
    34*x^2 + 68*x - 510
    Recall:
    a*x^2 + b*x + c
    x1 = ( - b + sqrt ( b*b - 4*a*c ) ) / ( 2*a)


    I have no idea how to go about doing this. Can someone help me out?


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  2. #2
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    Re: Find the Positive Root Of The Following Equation

    Divide both sides of the equation 34x^2 + 68x - 510 = 0 by 34.

    x^2 + 2x - 15 = 0


    By the quadratic formula, x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-15)}}{2(1)}

    =\frac{-2 \pm \sqrt{64}}{2} = \frac{-2 \pm 8}{2}

    The positive root is \frac{-2 + 8}{2} = 3.


    Alternatively, x^2 + 2x - 15 = 0 may be factored: (x+5)(x-3) = 0. The roots are -5 and 3.
    Last edited by richard1234; June 27th 2012 at 10:28 AM.
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  3. #3
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    Re: Find the Positive Root Of The Following Equation

    Quote Originally Posted by gregzaj1 View Post
    Positive root of the following equation:
    34*x^2 + 68*x - 510

    First, you understand that this is NOT an equation, don't you? I think you meant 34x^2+ 68x- 510= 0 which is an equation because it has "=" but I had to guess at the "0"

    Recall:
    a*x^2 + b*x + c
    Again, that is NOT an equation. You mean ax^2+ bx+ c= 0.
    Now, compare that with your 34x^2+ 68x- 510= 0. You should see that a= 34, b= 68, and c= -510

    x1 = ( - b + sqrt ( b*b - 4*a*c ) ) / ( 2*a)

    So "b*b- 4*a*c" is (68)(68)- 4(34)(-510)= 4624+ 69360= 73984 and the square root is 272.
    -b+sqrt(b*b- 4*a*c)= -68+ 272= 204. 2*a= 2*34= 68 so x= 204/68= 3.

    That wasn't so hard was it?

    (If you are very clever, like richard1234, you might recognise that 68= 2(34) and 510= 15(34) so dividing through by 34 at the start simplifies the numbers.)


    I have no idea how to go about doing this. Can someone help me out?

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