Find $k$ such that there are 2 solutions
I don't understand the working here. Specifically for when they consider -kx^2 +4kx = 0. I don't get how solving x from this allows us to determine what values will give us 2 solutions?
Can you instead factor the cubic in to a linear and quadratic term. Then use the discrimant for the quadratic to solve when it has 1 solution?
Edit: I can only post links to my image because I need 10+ reputation to do so. Upvote this and I'll edit it when I have enough. Cheers.