The sum of all the real roots of the equation [(e^2x - 4)(6e^2x - 5e^x + 1) = 0] is

Question

The sum of all the real roots of the equation $$({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0$$ is

Accepted Answer

$$({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0$$

Let ${e^x} = t$

$$\therefore$$ $({t^2} - 4)(6{t^2} - 5t + 1) = 0$

$$ \Rightarrow ({t^2} - 4)(2t - 1)(3t - 1) = 0$$

$$\therefore$$ t = 2, -2, $${1 \over 2}$$, $${1 \over 3}$$

$$\therefore$$ $${e^x} = 2 \Rightarrow x = \ln 2$$

${e^x} =  - 2$ (not possible)

$${e^x} = {1 \over 2} \Rightarrow x =  - \ln 2$$

$${e^x} = {1 \over 3} \Rightarrow x =  - \ln 3$$

$$\therefore$$ Sum of all real roots

$$ = \ln 2 - \ln 2 - \ln 3$$

$$ =  - \ln 3$$

Solutions · 1