Finding the minimum of $v^2 + x^2 - 2xv\gamma$ Let's say I have this expression $v^2 + x^2 - 2xv\gamma$, in which $\gamma$ is a constant. I want to find values for $v$ and $x$ to minimize the expression.
Sounds easy: just differentiate and set the result to zero. However, differentiating with respect to $v$ yields different results compared to differentiating w.r.t $x$.
$\partial/\partial v \rightarrow 2v -2x \gamma = 0 \rightarrow v = x\gamma$
$\partial/\partial x \rightarrow 2x -2v \gamma = 0 \rightarrow x = v\gamma$
Why is there a difference? Like, the first equation implies that this function hits an extremum when $v = x \gamma$. The value of the expression at that extremum is now $\gamma^2 x^2 + x^2 - 2x^2 \gamma^2 = x^2 - x^2 \gamma^2$. On the other hand, the second equation implies that this function hits an extremum when $v = x / \gamma$, and that the value at the extremum is actually $x^2/\gamma^2 - x^2$. The two expressions are plainly different (unless $\gamma = \pm 1$).
How do I interpret this discrepancy? What exactly are the two equations $v = x\gamma$ and $x = v \gamma$? What is the true value of the expression at the extremum?

$v=x\gamma$ and $x=v\gamma$ give $v=v\gamma^{2}$. If $\gamma \neq \pm 1$ we get $v=0$ and $x=v\gamma =0$. If $\gamma =\pm 1$ the expression becomes $(x+v)^{2}$ or $(x-v)^{2}$. Can you minimize this?
