Let $x=c$ be the common root of the polynomials CASE 1) $a=-2$ since $c^2-2c+b=c^2+c-2b\implies b=c\implies b^2-2b+b=b(b-1)=0\implies b=1$ which is indeed a solution CASE 2) $a\not= -2$ Summing up all the polynomials we get the polynomial $(a+2)(x^2+x+b)$ $c^2+c+b=0,c^2+c+ab=0\implies b=ab\implies a=1$ now for the polynomials to have only one common root we must have $b=\frac{1}{4}$ So : $(a,b)=(-2,1)-(1,\frac{1}{4})$

If $c$ is common root then $ac+b=c+ab \to (a-1)(c-b)=0$ Case 1: $a=1$ all polynomials are the same, so they have common root if $x^2+x+b$ has one root $\to b=\frac{1}{4}$ Case2: $a\neq 1,c=b: b^2+ab+b=0,ab^2+2b=0 \to b+a+1=0,ab+2=0 \to (a,b)=(-2,1)$