Thread: Find Vertical Line Intersecting Parabola Only Once

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.

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.

For any quadratic equation there is exactly one solution when the discriminant .

Originally Posted by Prove It

For any quadratic equation there is exactly one solution when the discriminant .

Thank you very much!!! I didn't realise that they worked out the discriminant. Once again, thanks!