jasperE3 10.04.2021 02:00 Find all pairs of real numbers $(x,y)$ satisfying $$(2x+1)^2+y^2+(y-2x)^2=\frac13.$$
jasperE3 10.04.2021 02:46 It expands as $24x^2-12xy+6y^2+12x+2=0$. Treating it as a quadratic in $x$, we get $y=\frac{3x\pm\sqrt{-3}(3x+1)}3$, and since $y\in\mathbb R$, we must have $x=-\frac13$, yielding the solution $\boxed{(x,y)=\left(-\frac13,-\frac13\right)}$.