If \(\alpha\) and \(\beta\) are the roots of \(x^2+7 x+3=0\) and \(\frac{2 \alpha}{3-4 \alpha}, \frac{2…
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 \(a+b+c=\)
11
0
243
81
Solution
\(\begin{aligned}
& \text {Let } \frac{2 \alpha}{3-4 \alpha}=y \Rightarrow 2 \alpha=3 y-4 \alpha y \\
& \Rightarrow \alpha(2+4 y)=3 y \Rightarrow \alpha=\frac{3 y}{2+4 y}
\end{aligned}\)
\(\because \alpha\) is root of quadratic equation \(x^2+7 x+3=0\),
So,
\(\begin{aligned}
& \qquad\left(\frac{3 y}{2+4 y}\right)^2+7\left(\frac{3 y}{2+4 y}\right)+3=0 \\
& \Rightarrow 9 y^2+84 y^2+42 y+48 y^2+48 y+12=0 \\
& \Rightarrow 141 y^2+90 y+12=0 \\
& \Rightarrow 47 y^2+30 y+4=0 \\
& \because y=\frac{2 \alpha}{3-4 \alpha} \text { is root of quadratic equation } \\
& a x^2+b x+c=0 . \\
& \therefore a=47, b=30 \text { and } c=4 \text { and } G C D \text { of } 47,30,4 \text { is } 1 . \\
& \therefore \quad a+b+c=47+30+4=81
\end{aligned}\)
Hence, option (4) is correct.