• # question_answer Sum of roots is -1 and sum of their reciprocal is $\frac{1}{6},$ then equation is A)  ${{x}^{2}}-6x+1=0$ B)  ${{x}^{2}}-x+6=0$ C)  $6{{x}^{2}}+x+1=0$ D)  ${{x}^{2}}+x-6=0$

Let the roots be $\alpha$ and $\beta$. Then, $\alpha +\beta =-1$ (given) and $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{1}{6}$ $\Rightarrow$ $\frac{\beta +\alpha }{\alpha \beta }=\frac{1}{6}$ $\therefore$ $-\frac{1}{\alpha \beta }=\frac{1}{6}$ $\Rightarrow$ $\alpha \beta =-6$ So, the required equation is ${{x}^{2}}+x-6=0$