# Quaratic Equation (roots and coefficients)

1. Oct 25, 2008

### ritwik06

1. The problem statement, all variables and given/known data
$$f(x)=x^{2}+ax+b$$
If v is one root of the quadratic then $$v^{2}-2$$ 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
$$-a=v^{2}+v-2$$
$$0=v^{2}+v+a-2$$
By putting discriminant >=0
I get:
1-4(a-2)>=0

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

2. Oct 26, 2008

### 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 $$\pm$$ 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.

3. Oct 30, 2008

### ritwik06

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!!!

4. Oct 30, 2008

### Staff: Mentor

Isn't this a description of the set of ordered pairs?
{(a, b): <the relation you found goes here>}

5. Oct 30, 2008

### ritwik06

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!!!

6. Oct 31, 2008

### 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.