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

Given equation is $9{\mathrm{x}}^2+6\mathrm{x}+1=0$

$\Rightarrow \mathrm{\alpha }+\mathrm{\beta }=\frac{-6}{9}=\frac{-2}{3}$ and $\mathrm{\alpha \beta }=1/9$

$\therefore \mathrm{\alpha }-\mathrm{\beta }=\sqrt{{(\mathrm{\alpha }+\mathrm{\beta })}^2-4\mathrm{\alpha \beta }}=\sqrt{\frac{4}{9}-4.\frac{1}{9}}=0$

$\Rightarrow \mathrm{\alpha }=\frac{-1}{3}, \mathrm{\beta }=\frac{-1}{3}$

$\therefore \mathrm{Equation} {\mathrm{x}}^2-(\mathrm{\alpha }+\mathrm{\beta })\mathrm{x}+\mathrm{\alpha \beta }=0$

$\Rightarrow {\mathrm{x}}^2+6\mathrm{x}+9=0$

(or)

$f\left(\frac{1}{x}\right)=0\Rightarrow x^2+6x+9=0$