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Math Help - Quadratic Equation Solutions

  1. #1
    Junior Member
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    Quadratic Equation Solutions

    Greetings,

    f(x)=x^2-(a^2+2a)x+a^3+a^2 is a given function.
    We are looking for all values of a such that at least one of the solutions of the function is within the interval [-1;1].

    So, I've come up with this:
    to find all relevent a values for exactly when one x is within [-1;1], we need this:
    -1 <= x <= 0, which happens when f(-1)f(1) <= 0. (1)

    To find all relevent values of a for when we need both solutions to be within [-1;1], we need:
    D>=0 (2)
    af(1)>=0 (3)
    -1<= -b/2a <= 1 (4)

    and we put it all from 1 to 4 in a system.

    For the solution of (1), I have:
    a is from (-1-sqrt5) / 2 to -1 U from (-1+sqrt5) / 2 to 1
    (Sorry, I couldn't get Latex to work.. )

    For (2), I believe all values of a do the job..

    For (3):
    a (-infinity;(-1-sqrt5)/2] U [1;+infinity]...

    and finally for (4):
    [-1-sqrt3;-1+sqrt3]

    So, when we combine all of this, eventually, I am left with:
    a [-1; (-1-sqrt5)/2]... which is not correct.

    The true thing is supposed to be:
    [(-1-sqrt5)/2; 1]

    Thank you..

    and sorry again for the lame writing...
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Logic View Post
    Greetings,

    f(x)=x^2-(a^2+2a)x+a^3+a^2 is a given function.
    We are looking for all values of a such that at least one of the solutions of the function is within the interval [-1;1].
    .
    Apply the quadratic formula to:

     x^2-(a^2+2a)x+a^3+a^2=0

    This gives the roots as a and a(a+1), so the first root is in [-1,1] when a is in this interval. So all we now need to do is determine those values of a for which a(a+1)is in [-1,1].

    RonL

    rONl
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