Question

If the quadratic equation $a{x}^2+bx+c=0$ with real coefficients has real and distinct roots in $(1,2)$ then $a$ and $5a+2b+c$

• A. have same sign.
• B. have opposite sign.
• C. are equal.
• D. are not real.

Answer 1

The correct option is A have same sign.

Let $\alpha ,\beta$ be the roots of $a{x}^2+bx+c=0$
Given that $1<\alpha ,\beta <2$


Case $\left(1\right):a>0$
$f\left(2\right)>0\Rightarrow 4a+2b+c>0\therefore 5a+2b+c>0$

Case $\left(2\right):a<0$
$f\left(2\right)<0\Rightarrow 4a+2b+c<0\therefore 5a+2b+c<0$

In both the cases $a$ and $5a+2b+c$ has same sign.

Alternate solution:
$a(5a+2b+c)=a^2(5+\frac{2b}{a}+\frac{c}{a})$
$=a^2[5-2(\alpha +\beta )+\alpha \beta ]$
$=a^2\left[\right(\alpha -2\left)\right(\beta -2)+1]>0[\because \alpha <2\text{and}\beta <2]$
$\therefore$ $a$ and $5a+2b+c$ have same sign.