Find the range of $k$ in $f(x) = \frac{x^2-k}{x-2}$

Question

I have the following question:

For real $x$, $f(x) = \frac{x^2-k}{x-2}$ can take any real value. Find the range of values $k$ can take.

Here is how I commenced:

$$ y(x-2) = x^2-k \\ -x^2 + xy - 2y + k = 0\\ $$

So we have $a=-1$, $b=y$, $c=(-2y+k)$. In order to find out the range of $f(x)$, i.e. the nature of its roots, we need the discriminant:

$$ y^2 - 4(-1)(-2y+k) \ge 0 \\ y^2 -8y + 4k \ge 0 $$

So $k$ must equal something to make that equation true - but the only way I understand "true" here is if I can factorise the equation after substituting a possible value of $k$. There are a few values which settle this criteria, like $k=4$, $k=3$ and $k=\frac{7}{4}$. The listed answer says that $k \ge 4$, which makes sense because the latter values produce $y$ values less than 0.

Is there a better way to find out the values of k in this situation? Or do I need to rely on factorising with trial and error?

Answer 1

Your last inequality was: $$y^2-8y+4k\ge0$$ For it to be true for all real values of y, the discriminant of the LHS should always be non-positive. So we have:$$64-16k\le0$$$$16k\ge64$$$$k\ge4$$

Answer 2

$$y^2 -8y + 4k = (y-4)^2+4(k-4)$$

For $y=4$, we need $k \geq 4$ to ensure that the discriminant is positive.