Question

The curve y= (k+2)x^2 meets the line y=(k+3)x +3. Find the range of possible values of k.

Answer 1

$(k+2)x^2=(k+3)x+3\\ (k+2)x^2-(k+3)x-3=0\\ \Delta\geq0 \wedge k\not=-2\\ \Delta=(-(k-3))^2-4\cdot(k+2)\cdot(-3)\\ \Delta=(-k+3)^2+12k+24\\ \Delta=k^2-6k+9+12k+24\\ \Delta=k^2+6k+33\\ k^2+6k+33\geq0\\ \Delta=6^2-4\cdot1\cdot33=36-132=-9$