Question

Cho đa thức P(x) = ax² + bx + c và 2a + b = 0. Chứng tỏ rằng P(-1). P(3) ≥ 0.

Answer 1

Ta có:

\(P\left(-1\right)=a\left(-1\right)^2+b\left(-1\right)+c\)

\(\Rightarrow P\left(-1\right)=a-b+c\)

\(P\left(3\right)=a.3^2+b.3+c\)

\(\Rightarrow P\left(3\right)=9a+3b+c\)

\(\Rightarrow P\left(3\right)-P\left(-1\right)=9a+3b+c-a+b-c\)

\(\Rightarrow P\left(3\right)-P\left(-1\right)=8a+4b\)

\(\Rightarrow P\left(3\right)-P\left(-1\right)=4\left(2a+b\right)\)

\(\Rightarrow P\left(3\right)-P\left(-1\right)=0\)

\(\Rightarrow P\left(3\right)=P\left(-1\right)\)

\(\Rightarrow P\left(-1\right).P\left(3\right)=P\left(3\right)^2\)

Vì \(P\left(3\right)^2\ge0\)

\(\Rightarrow P\left(-1\right).P\left(3\right)\ge0\)