Find the range of values of $k$ for which $kx^2 + 8x + k <6$ for all real values of $k$ 
Find the range of values of $k$ for which $kx^2 + 8x + k <6 $ for all real values of $k$.

I'm unsure if the discriminant must be greater than zero or less than zero.
My working steps: \begin{align}b^2 - 4ac = (8)^2 - 4(-2)(17-k) &> 0\\64 - 4(-2)(17-k) &> 0\\64 + 136 -8k  &> 0\\200 &> 8k\end{align} so my answer is $$k < 200/8.$$ 
 A: We have
$$kx^2+8x+k<6 \iff kx^2+8x+k-6<0$$
and this is always true when $k<0$ and
$$b^2-4ac=64-4k(k-6)<0 \implies k^2-6k-16>0$$
that is $k<-2$.
A: I suppose you want the range of values of $k$ for which $kx^2+8x+k<6$ for all $\color{red}x$.
It is equivalent to $\;kx^2+8x+k-6<0$ for all $x$. Now a quadratic polynomial has a constant sign if and only if it has no real root, i.e. if and only if its (reduced) discriminant $\Delta'=16-k(k-6)<0$. Furthermore, this sign is the sign of the leading coefficient $k$. Thus, you have to solve the system of inequations:
$$ k^2-6k-16>0,\qquad k<0. $$
Can you continue?
A: In order for a quadratic function to always be less than a constant, the leading coefficient must be negative.  We can then attempt to complete the square in such a way that the maximum value of the function is $6$, then apply the appropriate inequalities to $k$.
\begin{align}
kx^2+8x+k&=k(x^2+\frac8kx)+k\\
&=k(x^2+\frac8kx+\frac{16}{k^2}-\frac{16}{k^2})+k\\
&=k(x^2+\frac8kx+\frac{16}{k^2})-\frac{16}k+k\\
&=k(x+\frac4k)^2+k-\frac{16}k\end{align}
Therefore, we want $$k-\frac{16}k<6 \quad\text{AND}\quad k<0.$$
Solving the inequality, we get
\begin{align}
k-\frac{16}k&<6\\
k-6-\frac{16}k&<0\\
\frac{k^2-6k-16}{k}&<0\\
\frac{(k+2)(k-8)}{k}&<0\\
k&\in(-\infty,-2)\cup(0,8)
\end{align}
The largest intersection of the two inequations is $\boxed{k\in(-\infty,-2)}$, and that is the answer we want.
