Question

If $\alpha ,\beta$ are the roots of the equation $9x^2+6x+1=0$, 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)

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

$(x-\frac{1}{\alpha })(x-\frac{1}{\beta })=0$

$x^2-(\frac{1}{\alpha }+\frac{1}{\beta })x+\left(\frac{1}{\alpha \beta }\right)=0$

$x^2-\left(\frac{\alpha +\beta }{\alpha \beta }\right)x+\left(\frac{1}{\alpha \beta }\right)=0$

From the given equation we know

$\alpha +\beta =-\frac{6}{9}$

$\alpha \beta =\frac{1}{9}$

By substituting we get

$x^2-(-6)x+9=0$

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