# If $\alpha , \beta$ are the roots of the equation $9x^2+6x+1=0$, then the equation with the roots are $\frac{1}{\alpha}$, $\frac{1}{\beta}$, is real to :
( A ) $2x^2 + 3x + 18=0$
( B ) $x^2 + 6x + 9=0$
( C ) $x^2 + 6x -9=0$
( D ) $x^3 - 6x +9=0$