Originally Posted by
virussss123 I have a question with a solution to it but I can't seem to understand exactly how the solution is obtained. This question is related to Geometry.
The question is: Find all non-vertical lines that pass through the point Q(0,4) and have exactly one point in common with the Parabola y=x^2 + 7x + 20
Once the equation y=mx+4 is equated to y=x^2 +7x +20, the solution
x^2 + x(7-m) +16 = 0 is obtained.
Then it says: "This equation has exactly one solution when (7-m)^2 -4(16) = 0 and this gives: m-7=+-8, m=-1 or m=15
How do you get from x^2 + x(7-m) +16 = 0
to (7-m)^2 -4(16) = 0? Thank you.