Question

If $\alpha$ and $\beta$ are the roots of $x^{2}+7 x+3=0$ and $\frac{2 \alpha}{3-4 \alpha}, \frac{2 \beta}{3-4 \beta}$ are the roots of $a x^{2}+b x+c=0$ and $GCD$ of $a, b, c$ is $1$ , then $a+b+c=$

• A. 11
• B. 0
• C. 243
• D. 81

Answer 1

The Correct Option is D

Let $\frac{2 \alpha}{3-4 \alpha}=y$
$\Rightarrow 2 \alpha=3 y-4 \alpha y$
$\Rightarrow \alpha(2+4 y)=3 y$
$\Rightarrow \alpha=\frac{3 y}{2+4 y}$
$\because \alpha$ is root of quadratic equation
$x^{2}+7 x+3=0$,
So, $\left(\frac{3 y}{2+4 y}\right)^{2}+7\left(\frac{3 y}{2+4 y}\right)+3=0$
$\Rightarrow 9 y^{2}+84 y^{2}+42 y+48 y^{2}+48 y+12=0$
$\Rightarrow 141 y^{2}+90 y+12=0$
$\Rightarrow 47 y^{2}+30 y+4=0$
$\because y=\frac{2 \alpha}{3-4 \alpha}$ is root of quadratic equation
$a x^{2}+b x+ c=0 .$
$\therefore a=47, b=30$ and $c=4$ and $G C D$ of $47,30,4$ is $1 .$
$\therefore a +b +c=47+30+4=81$