For which a the matrix is positive-definite We have $A=\begin{pmatrix}a&2\\ 2&a+1\end{pmatrix}$
And I want to know for which a the matrix is positive-definite.
So our requirement is $det(A)\ge0$
$(a*(a+1))-4\ge0$
$a^2+a-4\ge0$
I understand that in the points $\frac{-1+\sqrt{17}}{2},\frac{-1-\sqrt{17}}{2}$ the function are equal to zero.
Is it true to say that because the function is equal to zero in the points so $ ∀a>\frac{-1+\sqrt{17}}{2},\frac{-1-\sqrt{17}}{2}$
$a^2+a-4\ge0$?

The last step is to note that since $a^2+a-4=(a-(-1+\sqrt{17})/2)(a-(-1-\sqrt{17})/2)$, we have that it is positive precisely on $(-\infty,(-1-\sqrt{17})/2)\cup((-1+\sqrt{17})/2,\infty)$.
