Question

Cho đa thức f(x)=ax\(^2\)+bx+c. Chứng minh rằng f(-2).f(3)≤ 0 nếu 13a+b+2c=0

Answer 1

Do \(13a+b+2c=0\Rightarrow\left(4a-2b+c\right)+\left(9a+3b+c\right)=0\)

\(\Rightarrow4a-2b+c=-\left(9a+3b+c\right)\)

Do đó:

\(f\left(-2\right).f\left(3\right)=\left(4a-2b+c\right)\left(9a+3b+c\right)\)

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

\(=-\left(9a+3b+c\right)^2\le0\) (đpcm)

Answer 2

\(f\left(-2\right)\cdot f\left(3\right)\)

\(=\left(4a-2b+c\right)\left(9a+3b+c\right)\)

\(=\left(13a+b+2c-9a-3b-c\right)\left(9a+3b+c\right)\)

\(=\left(-9a-3b-c\right)\left(9a+3b+c\right)=-\left(9a+3b+c\right)^2< =0\)