Explain how the discriminant can be used to show that the line 4x + 4y = 4 and the parabola y = 2x^2 - 5x + 6?

Question

...do not intersect.

The discriminants are 16 for the former and -23 for the latter (discriminant is b^2 - 4ac).

My friend said that since the parabola only had an imaginary solution, that it did not cross the x axis, but that doesn't mean they don't intersect.

Answer 1

You can learn math well, if you may follow the definitions and concepts well.
The definition of "discriminant" is b^2 - 4ac in a quadratic equation ax^2 + bx + c = 0. Your argument is not valid since neither the line 4x + 4y = 4, nor the parabola y = 2x^2 - 5x + 6, is a quadratic equation.
The parabola can be plotted as a curve in the x-y plan, such that every point is REAL. The line 4x + 4y = 4 can also be plotted in the x-y plan, such that every point is REAL. If they have a cross-section, the cross-section point/s must be REAL, since the cross-section point/s must be on both curve and the line.
Since the cross-section must satisfy both equations, we may put: y = 1 - x (the line) into y = 2x^2 - 5x + 6 (the parabola), and get:
1 - x = 2x^2 - 5x + 6, or:
2x^2 - 4x + 5 = 0
This equation will determine the x-coordinate/s of any cross-section/s. Now this is a quadratic equation, and thus has a "discriminant" = 16 - 40 = -24. Therefore this quadratic equation does not result any real x and thus the given line and curve does not have any cross-section/s.