Find the range of values of $k$ such that $kx^2+8x+3k<2$ for all real values of $x$ 
Find the range of values of k such that $kx^2+8x+3k<2$ for all real values of x

I found two answers, $k\le\frac 83$ or $k\le-2$
However, the teacher told me the answer is $k\le-2$
Why?
I hope someone can solve my problem, your solution is appreciated.

 A: The first thing that comes to my mind after seeing the inequality is to make the RHS $0$. So, we get the inequality
$$kx^2 + 8x + 3k - 2 < 0$$
Let $f(x) = kx^2 + 8x + 3k - 2$
Since all the values that $f(x)$ would return for $f : X \rightarrow Y$ where $X , Y \in \mathbb{R}$ are less than $0$, the parabola that the function would form would lie below the $x$- axis (as stated by Crostul). 
Another observation that one makes is that if $k > 0$ then the shape of the parabola would be like this $\cup$. This would mean that the graph would cross the $x$- axis at some point (which should not happen since the function returns all values less than $0$). So, it must be noted that $k < 0$.
As the shape of the parabola is $\cap$ and it lies below the $x$- axis, it is clear that the function will have no real zeroes i.e. discriminant ($D$) must be less than $0$. 
$$8^2 - 4k(3k - 2) < 0$$
On simplifying further, we get 
$$(-3k + 8)(k + 2) < 0$$   
Since the product of both the factors is less than $0$ $\therefore -3k + 8 < 0$ or $k + 2 < 0$ ($\because$ product of two numbers is negative if one of it is negative).
So, we get 2 sets of values for $k$ $$k > \frac{8}{3}$$$$k < -2$$
Since we know that $k < 0$,
$\therefore$ the answer is only $k < -2$.
A: I don't know how you solved, so I can't know why you got it wrong, but in order to show that your solution is incorrect, let $k=0$, and for all  $\ x>0.25$ it won't hold op.
A: In your question the parabola $$kx^2 + 8x + 3k - 2 < 0$$ will be only true when the following two conditions are true 
First the discriminant be less than so that the parabola doesnt have any root which prevent it from giving it both positive and negative values . 
After solving that we get $$ x>8/3$$ and $$x<-2$$ 
For it to ensure that your output for the following parabola be always negative , your parabola must be downward facing one so that it only gives negative for all values if x . For that condition to be satisfied the coefficient of $$ x^2$$ .ie $$ k$$ in this case must be less than zero 
$$ k< 0 $$ 
Hence your answer that required $$ k< 8/3$$ would be eliminated and the answer would be $$ k<-2$$ 
