1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Quaratic Equation (roots and coefficients)
  1. Quaratic Equations (Replies: 1)

  2. Roots of an equation (Replies: 2)

  3. Root of equation (Replies: 1)

  4. Roots of equation (Replies: 22)

Loading...