Given: $2x^2+kx+2$, under what conditions is it positive definite?
Since it's a quadratic, we take the discriminant $\Delta = k^2 - 16$. For it to be definitive, we need $\Delta \geq 0 $, so we solve it and:
$k \geq -4$ or $k \geq 4$. For +ve x-values, we need that the roots are +ve, so:
$\alpha + \beta = +ve$ and (obviously) $\alpha \beta = +ve$; so: $\frac{-k}{2} \gt 0$ and $\frac{2}{2} > 0$.
What to do after this?
Quadratic function $y=ax^2+bx+c$ is positive definite if $a>0$ and $\Delta <0$ .
Since $a=2>0$ we have that $\Delta=k^2-16$ must be less than zero .
EDIT :
$$k^2-16<0$$
$$(k-4)(k+4)<0$$
Now , we have to cases :
$(k-4)<0$ and $(k+4)>0$
$(k-4)>0$ and $(k+4)<0$
The first case give us solution $k\in(-4,4)$ and the second case give us $k \in \emptyset$ . Hence , final result is union of these two solutions , and therefore $k\in(-4,4)$ .