amparvardi wrote: For a real number $t$ and positive real numbers $a,b$ we have \[2a^2-3abt+b^2=2a^2+abt-b^2=0\] Find $t.$ $2a^2-3abt+b^2=2a^2+abt-b^2$ $\implies$ $4abt=2b^2$ Plugging $abt=\frac {b^2}2$ in $2a^2+abt-b^2=0$, we get $4a^2=b^2$ and so $b=2a$ and $\boxed{t=1}$

we have $4abt=2b^2$ so, $2at=b$ or $ 2a^2-3a(2at)t+4a^2t^2=0$ or $2at^2=2a^2$ or $t^2=1$ since $t=\frac{b}{2a}>0$ we have $ t=1$ then see $t=1$ and the possible $a,b$ are $a=2b$ for $a\in \mathbb R^+$