Question

Cho đa thức P(x)= \(ax^2+bx+c\)   và 2a+b=0

CMR: P(-1).P(3)\(\ge\)0

Answer 1

Có: \(\hept{\begin{cases}P\left(-1\right)=a-b+c\\P\left(3\right)=9a+3b+c\end{cases}}\)

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

\(=\left(a-b+c\right)\left[4\left(2a+b\right)+a-b+c\right]\)

\(=\left(a-b+c\right)\left(a-b+c\right)\)

\(=\left(a+b-c\right)^2\ge0\left(ĐPCM\right)\)

Answer 2

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

\(P\left(3\right)=a3^2+3b+c=9a+3b+c\)

từ đó suy ra \(P\left(-1\right).P\left(3\right)=\left(a-b+c\right)\left(9a+3b+c\right)\)

\(=\left(a-b+c\right)\left[\left(8a+4b\right)+a-b+c\right]\)

\(=\left(a-b+c\right)\left[4\left(2a+b\right)+a-b+c\right]\)

\(=\left(a-b+c\right)\left(a-b+c\right)=\left(a-b+c\right)^2\ge\)(đpcm)