Thread: range of rational quadratic function

I want to show the range of y is R for all x. y=(2x-1)/(2x^2-4x+1)

I made a quadratic equation and used the fact that the discriminant >=0 for real x. It was solved to y>=0.5(i-1).

What is the meaning of $\displaystyle y \ge .5(i-1)$? How do you define inequalities in the complex plane?

We want to show that for $\displaystyle \frac{2x-1}{2x^2 - 4x + 1} = k$, no matter what our choice of k is, there is always a real solution for x. Equation becomes

$\displaystyle 2x - 1 = 2kx^2 - 4kx + k$

$\displaystyle 2kx^2 - (4k+2)x + (k+1) = 0$

The discriminant D is

$\displaystyle D = (4k+2)^2 - 4(2k)(k+1) = 8k^2 + 8k + 4$, which is positive for all real k.

That is the discriminant I got. I thought I had to factorise it. Is there nothing further to do?

I am finding I can't make replies using IE8. Is this a common problem with this forum?