6.122. Form a second-degree equation, one of whose roots is equal to the sum, and the other to the product of the roots of the equation $a x^{2}+b x+c=0$.
Solution.

Let $y^{2}+p y+q=0$ be the desired equation with roots $y_{1}=x_{1}+x_{2}$, $y_{2}=x_{1} \cdot x_{2}$. By Vieta's theorem, we have

$$
\left\{\begin{array} { l } 
{ y _ { 1 } = x _ { 1 } + x _ { 2 } = - \frac { b } { a } } \\
{ y _ { 2 } = x _ { 1 } \cdot x _ { 2 } = \frac { c } { a } }
\end{array} \quad \text { and } \left\{\begin{array}{l}
y_{1}+y_{2}=-p \\
y_{1} \cdot y_{2}=q
\end{array}\right.\right.
$$

From this, $p=-\left(y_{1}+y_{2}\right)=\frac{b}{a}-\frac{c}{a}=\frac{b-c}{a}, q=y_{1} y_{2}=-\frac{b c}{a^{2}}$. We obtain the equation

$$
y^{2}+\frac{b-c}{a} y-\frac{b c}{a^{2}}=0 \Leftrightarrow a^{2} y^{2}+a(b-c) y-b c=0
$$

Answer: $a^{2} y^{2}+a(b-c) y-b c=0$.