AMC 10 Daily Practice Round 2

Complete problem set with solutions and individual problem pages

Problem 17 Medium

The roots of equation {{x}^{3}}+a{{x}^{2}}+bx+c=0 with respect to x are a,b,c, respectively, while a,b,c are rational numbers that are not all 0. Determine a+b+c.

  • A.

    -8

  • B.

    -1

  • C.

    0

  • D.

    1

  • E.

    8

Answer:B

By Vieta's Formula on higher power,

\begin{cases}a=-\left( a+b+c \right) \\ b=ab+bc+ca \\ c=-abc \\ \end{cases}. As c=-abc, we have c=0 or ab=-1.

① If c=0, then \begin{cases}2a+b=0 \\ b=ab \\ \end{cases}. Solve to get \left( 0,0,0 \right) (not applicable) or \left( 1,-2,0 \right);

② If ab=-1, then \begin{cases}ab+b+c=0 \\ b=bc+ca-1 \\ \end{cases}. Solve to get \left( 1,-1,-1 \right).

Therefore, \left( a,b,c \right)=\left( 1,-2,0 \right) or \left( 1,-1,-1 \right).

(a,b,c)=(1,-2,0) or (a,b,c)=(1-1,-1). Each case gives the sum of -1.

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