If 2 and 6 are the roots of the equation $a x^2+b x+1=0$, then the quadratic equation, whose roots are $\frac{1}{2 a+b}$ and $\frac{1}{6 a+b}$, is :

Question

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Sum }=8=-\frac{b}{a} \ & 	ext { Product }=12=\frac{1}{a} \quad \Rightarrow a=\frac{1}{12} \ & b=-\frac{2}{3} \ & 2 a+b=\frac{2}{12}-\frac{2}{3}=-\frac{1}{2} \ & 6 a+b=\frac{6}{12}-\frac{2}{3}=-\frac{1}{6} \ & 	ext { sum }=-8 \ & P=12 \ & x^2+8 x+12=0\end{aligned}$

If 2 and 6 are the roots of the equation $a x^2+b x+1=0$, then the quadratic equation, whose roots are…

Solution