Izmeystev I.V.

Find all such pairs of quadratic trinomials $x^{2}+a x+b, x^{2}+c x+d$, where $a$ and $b$ are the roots of the second trinomial, and $c$ and $d$ are the roots of the first.
By Vieta's theorem $a=-(c+d), b=c d, c=-(a+b), d=a b$. Therefore, $a+b+c=a+c+d=0, b=c d, d=$ $a b$. Consequently, $b=d=b c=a b$.

If $b=0$, then $d=0, c=-a$. If $b \neq 0$, then $a=c=1, b=d=-2$.

## Answer

$x^{2}+a x, x^{2}-a x$ ( $a-$ any number); $x^{2}+x-2, x^{2}+x-2$.