JEE Main 2024 (Online) 4th April Morning ShiftQuestion 1
If $2$ and $6$ are the roots of the equation $ax^{2}+bx+1=0$, then the quadratic equation, whose roots are $\dfrac{1}{2a+b}$ and $\dfrac{1}{6a+b}$, is :
Quadratic Equation · Relation Between Roots and Coefficients
Solution
Concept. The roots $2,6$ give $a$ and $b$; then the new roots are numbers. Formulas used. (1) For $ax^2+bx+c=0$: sum $=-\dfrac ba$, product $=\dfrac ca$Step 1 - $a$ and $b$. By (1), $12=\frac1a$ and $8=-\frac ba$, so $a=\frac1{12}$, $b=-\frac23$. Step 2 - New roots. $2a+b=\frac16-\frac23=-\frac12$ and $6a+b=\frac12-\frac23=-\frac16$. Their reciprocals are $-2$ and $-6$. Step 3 - The equation. Sum $-8$, product $12$: $x^2+8x+12=0$. Answer: (A)