Suppose that the equation $a x^2+b x+c=0$ has roots $\alpha$ and $\beta$, both of which are different from…
Suppose that the equation $a x^2+b x+c=0$ has roots $\alpha$ and $\beta$, both of which are different from $\frac{1}{3}$, then an equation whose roots are $\frac{1}{3 \alpha-1}$ and $\frac{1}{3 \beta-1}$ is
$(a+3 b+9 c) x^2+(3 b+2 a) x+a=0$
$(a+3 b+9 c) x^2-(3 b+2 a) x+a=0$
$(a+3 b+9 c) x^2+(3 b-2 a) x+a=0$
$(a+3 b+9 c) x^2-(3 b-2 a) x+a=0$
Solution
$a x^2+b x+c=0...(i)$
Let $\alpha$ and $\beta$ are the roots of Eq. (i).
$
\alpha+\beta=-\frac{b}{a} \text { and } \alpha \beta=\frac{c}{a}
$
Now, the quadratic equation whose roots are
$
\begin{aligned}
& \frac{1}{3 \alpha-1}, \frac{1}{3 \beta-1} \\
& \text { Sum of roots }=\frac{1}{3 \alpha-1}+\frac{1}{3 \beta-1} \\
& =\frac{3 \beta-1+3 \alpha-1}{(3 \alpha-1)(3 \beta-1)}=\frac{3(\alpha+\beta)-2}{9 \alpha \beta-3(\alpha+\beta)+1} \\
& =\frac{3\left(-\frac{b}{a}\right)-2}{9\left(\frac{c}{a}\right)-3\left(-\frac{b}{a}\right)+1} \\
& =\frac{-3 b-2 a}{9 c+3 b+a}=-\frac{(3 b+2 a)}{a+3 b+9 c} \\
&
\end{aligned}
$
Product of roots $=\frac{1}{(3 \alpha-1)(3 \beta-1)}$
$
\begin{aligned}
& =\frac{1}{9 \alpha \beta-3(\alpha+\beta)+1} \\
& =\frac{1}{9\left(\frac{c}{a}\right)-3\left(-\frac{b}{a}\right)+1} \\
& =\frac{a}{a+3 b+9 c}
\end{aligned}
$
Now, the required equation
$
x^2-\left\{-\left(\frac{3 b+2 a}{a+3 b+9 c}\right) x\right\}+\frac{a}{a+3 b+9 c}=0
$
$\left[\because\right.$ for any quadratic equation; $\therefore x^2-$ (sum of roots $) x+$ Product of roots $=0]$
$
\Rightarrow \quad(a+3 b+9 c) x^2+(3 b+2 a) x+a=0
$