4(1132). For what values of $a$ do the quadratic trinomials $x^{2}+$ $+a x+1$ and $x^{2}+x+a$ have a common root?
Solution. Let $x=x_{1}$ be the common root of the given quadratic trinomials. Then $x_{1}^{2}+a x_{1}+1=0$ and $x_{1}^{2}+x_{1}+a=0$, i.e., $x_{1}^{2}+a x_{1}+1=x_{1}^{2}+x_{1}+a, \quad a \cdot\left(x_{1}-1\right)=x_{1}-1, \quad\left(x_{1}-1\right) \cdot(a-1)=0$. From this, $a=1$ or $x_{1}=1$.

When $a=1$, each of the given quadratic trinomials has the form $x^{2}+x+1$, which has no solutions in the set of real numbers. Therefore, the given quadratic trinomials have a common root $x=1$. Find the value of $a$ for which $x=1$. When $x=1$, we have $1+a+1=0$, from which $a=-2$.

Answer: $a=-2$.

Remark. It is advisable to do a check. When $a=-2$, we have the quadratic trinomials $x^{2}-2 x+1$ and $x^{2}+x-2$, which have a common root $x=1$.