This appeared as a subproblem in a linear algebra exercise.
What values can $γ_{ij} \in \Bbb R,i=1,2,j=1,2$, have such that $$a_1^2γ_{11}+a_1a_2(γ_{12}+γ_{21})+a_2^2γ_{22} \ge 0?$$
where $a_k \in \Bbb R,k=1,2$
This is closely related to the $Δ$ of a polynomial $ax^2+bx+c$, but since it is not a polynomial we have here, I can't set the $Δ \ge 0$. So how do we approach such a problem and find $γ_{ij}$?
You've almost got it! The key realization is that the expression $$ a_1^2γ_{11}+a_1a_2(γ_{12}+γ_{21})+a_2^2γ_{22} = a_2^2 \bigg( \Big(\frac{a_1}{a_2}\Big)^2γ_{11}+\frac{a_1}{a_2}(γ_{12}+γ_{21})+γ_{22} \bigg) $$ is essentially a quadratic polynomial, but in the variable $\frac{a_1}{a_2}$ (or its reciprocal, alternatively). So the discriminant is indeed the thing you want to look at (as well as the signs of $\gamma_{11}$ and $\gamma_{22}$, of course).