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.
Best Answer
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 $k < -2$.