JEE Main 2022 (Online) 24th June Evening ShiftQuestion 1
The sum of all the real roots of the equation $\left(e^{2x}-4\right)\left(6e^{2x}-5e^{x}+1\right)=0$ is
Quadratic Equation · Equations Reducible to Quadratic
Solution
Concept. Each bracket is a quadratic in $t=e^x>0$. Formulas used. (1) $e^x=t>0\iff x=\ln t$Step 1 - First bracket. $e^{2x}=4$, so $x=\ln2$. Step 2 - Second bracket. $6t^2-5t+1=(2t-1)(3t-1)=0$: $t=\frac12,\frac13$, so $x=-\ln2,\,-\ln3$. Step 3 - Sum. $\ln2-\ln2-\ln3=-\ln3$. Answer: (B)