I'm trying to find the range of values $k$ can take given that, for real $x$, $f(x) = \frac{x^2+3k}{x+k}$ can take any real value. These are the steps I've taken so far:
$$ xy + ky - x^2 - 3x = 0 $$
Which gives us $a=-1$, $b=(y-3)$, $c=yk$. Using the discriminant:
$$ (y-3)^2 - 4(-1)(yk) \ge 0 \\ y^2 + y(-6 + 4k) + 9 \ge 0 $$
Taking the discriminant again, we have $a=1$, $b=(4k-6)$, $c=9$ which gives us:
$$ (4k-6)^2 - 4(9) \ge 0 \\ k(k-3) \ge 0 $$
So using the table method:
$$ \begin{array}{c|lcr} \space & k \le 0 & 0 \le k \le 3 & k \ge 3 \\ k & \text{-ve} & \text{+ve} & \text{+ve} \\ k-3 & \text{-ve} & \text{-ve} & \text{+ve} \\ k(k-3) & \text{+ve} & \text{-ve} & \text{+ve} \\ \end{array} $$
So it clearly seems that the range is $k \le 0, k\ge3$ - but the listed answer is $0 \le k \le 3$, which doesn't make sense because that results in a negative value and invalidates the inequality equation. Where am I going wrong?
After getting $$y^2+(-6+4k)y+9\ge 0,$$ what you need is $$(-6+4k)^2-4\cdot 1\cdot 9\color{red}{\le} 0$$ instead of $$(-6+4k)^2-4\cdot 1\cdot 9\ge 0.$$