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 $x$.

Question

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!!

Answer 1

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 $k$.

Answer 2

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 $k\in (-\infty,k_1)\cup(k_2,\infty)$.

Answer 3

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.