1. ## intersection

the question is:
the parabola with equation $y = 2x^2 + 4x - 8$ intersects the line $y + kx + 4k = 0$ where k is a constant.
show that any points of intersection have x coordinates that satisfy $2x^2 + (k + 4)x + 4(k - 2) = 0$

how would you show this? thanks

2. Originally Posted by mark
the question is:
the parabola with equation $y = 2x^2 + 4x - 8$ intersects the line $y + kx + 4k = 0$ where k is a constant.
show that any points of intersection have x coordinates that satisfy $2x^2 + (k + 4)x + 4(k - 2) = 0$

how would you show this? thanks
when the two curves intersect they have the same x,y cordinates so

$y=2x^2+4x-8$

$y=-kx-4k$

$-kx-4k = 2x^2 +4x-8$

$2x^2 +4x+kx-8+4k=0$

take factors

$2x^2 +x(k+4) + 4(k-2)=0$