# Find the range of values $k$ can take given that, for real $x$, $f(x) = \frac{x^2+3k}{x+k}$

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 $$(-6+4k)^2-4\cdot 1\cdot 9\ge 0.$$
• @hohner: Do you understand the relation between the sign of the discriminant and the number of real roots of $ax^2+bx+c=0$? Oct 4, 2014 at 17:08
• If the discriminant is $>0$, then there are 2 distinct real roots. If it's less than 0, there are no real roots (but 2 complex roots). Oct 4, 2014 at 17:10
• @hohner: Right. Then, consider the parabola $Y=y^2+(-6+4k)y+9$. We want the range of $k$ such that $Y=y^2+(-6+4k)y+9\ge 0$ for all $y$. So, $y^2+(-6+4k)y+9=0$ must not have two real roots (having one real root is fine, which means the parabola is tangent to $y$-axis), so we need $D\le 0$ instead of $D\ge 0$. Oct 4, 2014 at 17:14