Find the value of $k$ for which the inequality $x^2-2(4k-1)x+15k^2-2k-7>0$ is valid for ANY value of $x$. I think you have to use the determinant $D=b^2-4ac$ to find out the value of $k$.
But I just can't use it properly.
Please help.
Thanks!!
 A: Guide:
This is a convex quadratic function, to be positive everywhere, it doesn't intersect with the $x$-axis. 
Hence set $D<0$ and solve for $k$.
A: I assume you know how to solve quadratic equations and know what $\Delta = b^2-4ac$ is. Note that the parabola given by your lhs. expression has a positive leading coefficient. This means it has a minimum value and goes over to $+\infty$ with $|x|$. Now, you want to find the intersection points with the abscissa axis. There are three cases: 
$\Delta>0$ and there are 2 real values $x_0,x_1$ in between which the function is negative, 
$\Delta=0$ and there is only one intersection point $x_\ast$
$\Delta<0$ and there are no real intersection points, thus the function is strictly positive everywhere.
You are interested in the third case, so you need to solve $\Delta<0$ in terms of $k$. Note that, this inequality itself involves a quadratic expression of $k$, so you need to apply the same reasoning. This will probably give you an answer of the form $k\in (-\infty,k_1)\cup(k_2,\infty)$.
A: Note that $x^2-2(4k-1)x+15k^2-2k-7\gt0$ if and only if
$$(x-(4k-1))^2=x^2-2(4k-1)x+16k^2-8k+1\gt k^2-6k+8=(k-3)^2-1$$
The extreme left hand side takes the value $0$ at $x=4k-1$ and is otherwise positive.  The extreme right hand side is negative for $2\lt k\lt4$ and non-negative otherwise.  Thus the inequality $x^2-2(4k-1)x+15k^2-2k-7\gt0$ holds for all $x$ (in particular for $x=4k-1$) if and only if $2\lt k\lt4$. If $k$ is required to be an integer, then $k=3$ is the only answer.
