Question

If the equation $a{x}^2+bx+c=0$ has two positive and real roots, then the equation $3a{x}^2+(2a+3b)x+b+3c=0$ has

• A. At least one positive real root
• B. Two positive real roots
• C. At least one negative real root
• D. No real root

Answer 1

The correct option is A At least one positive real root

Consider $f\left(x\right)=e^{3x}(a{x}^2+bx+c).$
$f\left(x\right)=0$ has two positive real roots. Using Rolle's theorem, We can say $f^{\prime }\left(x\right)=0$ has at least one real root between two roots of $f\left(x\right)=0.$
Now, $f^{\prime }\left(x\right)=e^{3x}(2ax+b)+3e^{3x}(a{x}^2+bx+c)$
$\Rightarrow e^{3x}(3a{x}^2+(2a+3b)x+b+3c)=0$ has at least one positive root.
$\therefore 3a{x}^2+(2a+3b)x+b+3c=0,(\because e^{3x}>0)$
has at least one positive root.