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:

The Discriminant of a Quadratic Equation – Part 1
The Discriminant of a Quadratic Equation – Part 2
The Discriminant of a Quadratic Equation – Part 3
The Discriminant of a Quadratic Equation – Part 4

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 d, e, and f are integers. For which of the following will there be more than one (x, y) solution, with real-number coordinates for the system?

A. e² + 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 x² in 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 a = e, b = d, 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. 

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