If alpha and beta are the roots of 9x^2 + 6x | vedclass

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(C) Given the quadratic equation $9x^2 + 6x + 1 = 0$.Let the roots be $\alpha$ and $\beta$. The sum of roots is $\alpha + \beta = -\frac{6}{9} = -\fra

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$x^2 + 6x + 9 = 0$

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(C) Given the quadratic equation $9x^2 + 6x + 1 = 0$.Let the roots be $\alpha$ and $\beta$. The sum of roots is $\alpha + \beta = -\frac{6}{9} = -\frac{2}{3}$ and the product of roots is $\alpha\beta = \frac{1}{9}$.We need to find the equation whose roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.The sum of the new roots is $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-2/3}{1/9} = -\frac{2}{3} 	imes 9 = -6$.The product of the new roots is $\frac{1}{\alpha} 	imes \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{1/9} = 9$.The required quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Substituting the values,we get $x^2 - (-6)x + 9 = 0$,which simplifies to $x^2 + 6x + 9 = 0$.

If $\alpha$ and $\beta$ are the roots of $9x^2 + 6x + 1 = 0$,then the equation with the roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is

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