Question

If roots of equation $a{x}^2+bx+c=0$, are distinct and less than unity, then which of the following is/are true?

• A. $\frac{9a+6b+4c}{25a+10b+4c}$ is positive
• B. $\frac{9a+3b+c}{a+b+c}$ is positve
• C. $\frac{9a+6b+4c}{25a+10+4c}$ is negative
• D. $\frac{c}{4a+2b+c}$ is negative

Answer 1

Answer: (A), (B)

$\frac{9a+3b+c}{a+b+c}$ is positve


$\frac{f(\frac{3}{2})}{f\left(\frac{5}{2}\right)}>0\Rightarrow \frac{9a+6b+4c}{25a+10b+4c}$ is positive & $\frac{f\left(3\right)}{f\left(1\right)}>0\Rightarrow \frac{9a+3b+c}{a+b+c}$ is positive.

We cannot comment on $\frac{f\left(0\right)}{f\left(2\right)}$.

Same in downward parabola.