2(1−k)x2 −4kx+k=0
Solve for x ?

Question

2(1−k)x2 −4kx+k=0
Solve for x ?

anonymous · Answer

sorry, not written properly :
2(1−k)x^2 −4kx+k=0

mimi_x3 · Answer

is this pertained to discriminants?

anonymous · Answer

Just to use quadratic formula, but don't know how to get it into ax^2+bx+c form c:

anonymous · Answer

It's in that form

mimi_x3 · Answer

i think it's related to this thingy
D = b^2 -4ac

jhannybean · Answer

$$2x^2 -2kx^2 -4kx+k=0$$$$x^2-kx^2 -2kx +\frac{k}{2}=0$$

anonymous · Answer

a = 2(1-k)

jhannybean · Answer

It is, mimi, but it had to be reduced first, lol

jhannybean · Answer

You can also use the completing the square method here.

mimi_x3 · Answer

lel my bad...it has been almost 4 years since ive touched this

jhannybean · Answer

Mmhmm.

anonymous · Answer

I'm sorry, I still don't get it! What exactly is a, b, and c and how did you get it ?

jhannybean · Answer

Vertex form of a quadratic: $$y=\color{red}ax^2+\color{red}bx+\color{red}c$$your equation:$$y=\color{red}{2(1-k)}x^2\color{red}{-4k}x+\color{red}k=0$$ Now just match up a,b,c from the quadratic to the variables in your equation, and then apply the quadratic formula, $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$Guess that would be the easiest way of going about it.

jhannybean · Answer

sorry, it's not the vertex form of the quadratic, forgot to erase that. It's just the parabolic form of the quadratic.