range of $k$ in quadratic equation the exhaustive set of values of $k$ for which  $|kx-2|=2x^2+kx+4$ has at least one positive root 
solution i try $|kx-2|$ is either $(kx-2)$ or $-(kx-2)$
$kx-2=2x^2+kx+4\Rightarrow 2x^2=-6$ no real solution
$-kx+2=2x^2+kx+4\Rightarrow x^2+kx+4=0$
$\displaystyle x=\frac{-k\pm \sqrt{k^2-16}}{4}$
$|k|\geq 4$ i did not understand how i use condition at least one positive root

If $-kx + 2 = 2x^2 + kx + 4$ then $2 = 2x^2 + 2kx + 4$ thus $0 = 2x^2 + 2kx - 2$. $$\therefore x^2 + kx - 1 = 0\tag1$$ and not $$x^2 + kx + 4 = 0.\tag2$$ Equation $(1)$ is correct, but Eq. $(2)$ is false. Therefore, you should have that $$x = \frac{-k\pm\sqrt{k^2 - 4}}{2}$$ which brings us to the fact that $|k|\geq 2$. Since we need a positive root $x$, then for some $n\in\mathbb{Z}$, we have $k^2 - 4 = n^2$. Since $4 = 2^2$ then we have that $k = 2$. $$\therefore x = \frac{-2\pm 0}{2} = -1.$$ This means that $k\neq 2$ because in this case, $x<0$. But, remember that $2^2 = (-2)^2$ so that means we can let $k = -2$. $$\therefore x = \frac{2\pm 0}{2} = 1.$$ It follows, then, that $(k, x) = (-2, 1)$ and $x$ has at least one positve solution.
