This equation has no solution when the discriminant is negative
Note: This is from a Vector Calculus course, so I hope it belongs in this section.
The question:
For what values of 'a' do the surfaces and not intersect?
My attempt:
I rearranged the first equation like so:
Then equated it with the other:
Usually I'd solve for z to find the conditions for the intersection, but we have two variables now.
I then solved for a:
So when z = -1, 'a' isn't defined. I then substituted this value into the very first equation to see what happens, but I just get the origin point.
I don't think I'm doing this correctly. Any assistance would be great!
This is the "wrong way". Solve the equation for z, not a. That is a quadratic equation for z so there will be no real number solution for z if the "discriminant" ( ) is negative.
So when z = -1, 'a' isn't defined. I then substituted this value into the very first equation to see what happens, but I just get the origin point.
I don't think I'm doing this correctly. Any assistance would be great!