If 4a+2b+c=0, then the equation 3ax2+2bx+c=0 has at least one real root lying in the interval

# If $4a+2b+c=0,$ then the equation $3a{x}^{2}+2bx+c=0$ has at least one real root lying in the interval

1. A

2. B

3. C

4. D

none of these

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### Solution:

Let $f\left(x\right)=a{x}^{3}+b{x}^{2}+cx.$Then, $f\left(0\right)=0$ and

$f\left(2\right)=2\left(4a+2b+c\right)=0$. Therefore, 0 and 2 are roots of $f\left(x\right).$ Therefore, ${f}^{\mathrm{\prime }}\left(x\right)=0$ i.e. $3a{x}^{2}+2bx+c=0$ has a root between 0 and 2.

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