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Homework Help: Quaratic Equation (roots and coefficients)

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]f(x)=x^{2}+ax+b[/tex]
    If v is one root of the quadratic then [tex]v^{2}-2[/tex] is the second root. Find all possible ordered pairs (a,b). Universal set is the set of real numbers.


    3. The attempt at a solution
    [tex]-a=v^{2}+v-2[/tex]
    [tex]0=v^{2}+v+a-2[/tex]
    By putting discriminant >=0
    I get:
    1-4(a-2)>=0

    But in the case of product of roots I get a cubic!!!

    Please help me with this. How can I get th exact ordered pairs??
     
  2. jcsd
  3. Oct 26, 2008 #2

    Mark44

    Staff: Mentor

    I think what you need to do here is to look at the graph of f(x) = x^2 + ax + b.
    The graph is a parabola that opens upward. The x-intercepts (roots) are the solutions of x^2 + ax + b = 0, and are x = (-a [tex]\pm[/tex] sqrt(a^2 - 4b)) / 2.

    For there to be two distinct solutions, what has to be true about the discriminant? From that you get a set of ordered pairs (a, b). That's what this problem seems to be asking for.
     
  4. Oct 30, 2008 #3
    The answer to that is that the discriminant need to be positive. It gives me a relation between a and b, not the ordered pairs!!!
     
  5. Oct 30, 2008 #4

    Mark44

    Staff: Mentor

    Isn't this a description of the set of ordered pairs?
    {(a, b): <the relation you found goes here>}
     
  6. Oct 30, 2008 #5
    Re: Quadratic Equation (roots and coefficients)

    Nope! I also thought that but the answer given at the back of my book lists the exact ordered pairs!!!
     
  7. Oct 31, 2008 #6

    Mark44

    Staff: Mentor

    By completing the square, we can rewrite f(x) = x^2 +ax + a^2/4 + b - a^2/4.
    Or, f(x) = (x + a/2)^2 + b - a^2/4.

    This parabola opens up and has a vertex somewhere along the line x = -a/2.

    We can choose a = -2 to put the vertex on the line x = -1, and if we choose b with care, we can run the parabola through (0,0) and (-2, 0). That way, we can let v = 0, so v^2 = 0, and v^2 - 2 = 0. We will have found a and b so that v is a root and v^2 - 2 is a root.

    I've given you a; all you have to do is to find b.
     
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