### The Discriminant of a Quadratic Equation Part 5

Today I’d like to solve the problem about discriminants from the last post. You can review parts 1, 2, 3, and 4 of our discussion on the discriminant here:

Last time, I asked you to solve the following problem:

Example: You are given the following system of equations.

dx + ey = f
y = x2

where ab, and c are integers. For which of the following will there be more than one (xy) solution, with real-number coordinates for the system?

Ae2 + 4df < 0
B. e2 – 4df < 0
C. d2 – 4ef < 0
D. e2 – 4df > 0
E. d2 + 4ef > 0

Make sure to try this problem yourself before reading the following solution.

Solution: By the second equation, we have y = x2, so we can replace y by xin the first equation to get

dx + ex2 = f

Writing this quadratic equation in general form gives the following.

ex2 + dx – f = 0

So, we have aebd, and c f . Thus, the discriminant is

Δ = b2 – 4ac = d2 – 4e( f) = d2 + 4ef

We want the quadratic equation to have two real solutions. Therefore, the discriminant must be positive. So, we must have

d2 + 4ef > 0

This is choice E