1. Given three quadratic trinomials: $:_{1}(x)=x^{2}+$ $b_{1} x+c_{1}, P_{2}(x)=x^{2}+b_{2} x+c_{2}, P_{3}(x)=$ $x^{2}+b_{3} x+c_{3}$. Then the equation $\left|P_{1}(x)\right|+P_{2}(x)=$ $\left|P_{3}(x)\right|$ has at most $(\quad)$ roots.
(A) 4
(B) 6
(C) 8
(D) None of the above
$-1 .(\mathrm{C})$
Each root of the original equation should be a root of a quadratic trinomial of the form $\pm P_{1}(x)+P_{2}(x) \pm P_{3}(x)$, and there are 4 such quadratic trinomials. Since the coefficient of the $x^{2}$ term has the form $\pm 1+1 \pm 1$, no matter how the signs are chosen, the coefficient of the $x^{2}$ term will not be 0. Furthermore, since $P_{2}(x)$ does not change sign, among these 4 quadratic trinomials, there are no two where each corresponding coefficient is the opposite of the other, meaning they are 4 different quadratic equations. Therefore, the number of roots will not exceed 8.