1. Let $D$ be the discriminant of the reduced quadratic trinomial $x^{2}+a x+b$. Find the roots of the trinomial, given that they are distinct and one of them is equal to $D$, and the other is equal to $2D$.
1. By Vieta's theorem $b=D \cdot 2D=2D^{2}, a=-(D+2D)=-3D$, i.e., the quadratic trinomial is equal to $x^{2}-3Dx+2D^{2}$. Its discriminant $D=(-3D)^{2}-4 \cdot 2D^{2}=D^{2}$, from which $D=D^{2}$, i.e., $D=0$ (in this case both roots are the same and equal to 0) or $D=1$ (in this case the roots are 1 and 2).