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(A) Given the equation $ax^2+bx+c=0$ with roots $\alpha$ and $\beta$. From the relation between roots and coefficients,we have $\alpha+\beta = -\frac{

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$(a+3b+9c)x^2+(3b+2a)x+a=0$

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(A) Given the equation $ax^2+bx+c=0$ with roots $\alpha$ and $\beta$. From the relation between roots and coefficients,we have $\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$. Let the new roots be $y_1 = \frac{1}{3\alpha-1}$ and $y_2 = \frac{1}{3\beta-1}$. The sum of the new roots is $S = \frac{1}{3\alpha-1} + \frac{1}{3\beta-1} = \frac{3\beta-1+3\alpha-1}{(3\alpha-1)(3\beta-1)} = \frac{3(\alpha+\beta)-2}{9\alpha\beta-3(\alpha+\beta)+1}$. Substituting the values,$S = \frac{3(-\frac{b}{a})-2}{9(\frac{c}{a})-3(-\frac{b}{a})+1} = \frac{-3b-2a}{9c+3b+a} = -\frac{3b+2a}{a+3b+9c}$. The product of the new roots is $P = \frac{1}{(3\alpha-1)(3\beta-1)} = \frac{1}{9\alpha\beta-3(\alpha+\beta)+1} = \frac{1}{9(\frac{c}{a})-3(-\frac{b}{a})+1} = \frac{a}{a+3b+9c}$. The required quadratic equation is $x^2 - Sx + P = 0$,which gives $x^2 - (-\frac{3b+2a}{a+3b+9c})x + \frac{a}{a+3b+9c} = 0$. Multiplying by $(a+3b+9c)$,we get $(a+3b+9c)x^2 + (3b+2a)x + a = 0$.

Suppose that the equation $ax^2+bx+c=0$ 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

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