Question

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

• A. $2x^2+3x+18=0$
• B. $x^2+6x-9=0$
• C. $x^2+6x+9=0$
• D. $x^2-6x+9=0$

Answer 1

Answer: (C)

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

Equation with roots having reciprocals of given
equation is $f\left(\frac{1}{\alpha }\right)=0$
$\Rightarrow 9(\frac{1}{\alpha }{)}^2+6(\frac{1}{\alpha })+1=0$
$\Rightarrow 9+6\alpha +{\alpha }^2=0$
$\therefore x^2+6x+9=0$ is the required equation