Question

If $\alpha ,\beta$ are the roots of the equation $m{x}^2+6x+(2m-1)=0\forall m\in R-\{0,\frac{1}{2}\},$ then the quadratic equation with roots as $\frac{1}{\alpha },\frac{1}{\beta }$ is:

• A. $(2m-1)x^2-6x+m=0$
• B. $m{x}^2+(2m-1)x+6=0$
• C. $(2m-1)x^2+6x+m=0$
• D. $6x^2-(2m-1)x-m=0$

Answer 1

The correct option is C $(2m-1)x^2+6x+m=0$

Given: $\alpha ,\beta$ are the roots of $m{x}^2+6x+(2m-1)=0\forall m\in R-\{0,\frac{1}{2}\}$
$\Rightarrow$ Sum of Roots $\alpha +\beta =\frac{-6}{m}\cdots \left(i\right)$
And similarly Product of Roots $\alpha \beta =\frac{2m-1}{m}\cdots \left(ii\right)$

To find: Quadratic Equation with Roots as $\frac{1}{\alpha },\frac{1}{\beta }$

For this transformed Equation:
Sum of Roots $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{\frac{-6}{m}}{\frac{2m-1}{m}}=\frac{-6}{2m-1}\cdots \left(iii\right)$
Similarly, Product of Roots
$\frac{1}{\alpha }\cdot \frac{1}{\beta }=\frac{m}{2m-1}\cdots \left(iv\right)$
Thus, the transformed quadratic equation will be:
$x^2-(\frac{1}{\alpha }+\frac{1}{\beta })x+\frac{1}{\alpha \beta }=0\Rightarrow x^2-(\frac{-6}{2m-1})x+\frac{m}{2m-1}=0\Rightarrow (2m-1)x^2+6x+m=0$
Which is the required quadratic equation.