Necessary and sufficient condition for quadratic forms

Question

Consider the quadratic form $Q(u,v)=au^2 + 2buv +cv^2$. Upon completing the square we will obtain $$Q(u,v)=a\left(u+\frac{bv}{a}\right)^2+\left(c-\frac{b^2}{a}\right)v^2,$$ we assume that $a\ne 0$.

There is a theorem which says that

A set of necessary and sufficient conditions that $Q(u,v)$ be positive for all nontrivial $u$ and $v$ is that $$a>0, \text{ } \left|\begin{matrix} a & b\\ b & c \end{matrix}\right|>0$$

However, what if $a>0$ but $\left(c-\frac{b^2}{a}\right)<0$? Then either $c$ is negative or $c<\frac{b^2}{a}$. Why wouldn't this satisfy $Q(u,v)$ as being positive?

This is also equivalent to

$$ac-b^2>0 \text{ (1)}$$ $$ac-b^2<0 \text{ (2)}$$

with (1) implying that $c>0$ whenever $a>0$ but $ac$ can also be positive when $a<0$ and $c<0$ and $ac>b^2$. Likewise, (2) can also be negative when $a,c>0$ and $ac<b^2$.

What is it that I'm missing here?

Answer 1

when $a > 0$ but $c - \frac {b^2}{a} < 0,$ test with pair $$ u = -b, \; \; v = a . $$ Then $$ a u^2 + 2 b u v + c v^2 = a b^2 - 2 b b a + c a^2 = ca^2 - a b^2 = a(ca-b^2) < 0 $$

Answer 2

Note indeed that the condition

$$\left(c-\frac{b^2}{a}\right)<0 \iff ac-b^2<0$$

is the opposite of

$$\left|\begin{matrix} a & b\\ b & c \end{matrix}\right|>0 \iff ac-b^2>0$$