The number of real roots of the equation $e^{4 x} - e^{3 x} - 4 e^{2 x} - e^{x} + 1 = 0$ is equal to

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Solution

Let $e^{x} = t, (t > 0)$
$t^{4} - t^{3} - 4 t^{2} - t + 1 = 0$
$\Rightarrow (t^{2} + \frac{1}{t^{2}}) - (t + \frac{1}{t}) - 4 = 0 \Rightarrow {(t + \frac{1}{t})}^{2} - (t + \frac{1}{t}) - 6 = 0$
Let $t + \frac{1}{t} = u (u > 2)$
$(u - 3) (u + 2) = 0$
$\Rightarrow u = 3, - 2$ (rejected)
For $u = 3$
$\Rightarrow t + \frac{1}{t} = 3 \Rightarrow t^{2} - 3 t + 1 = 0$
$t = \frac{3 \pm \sqrt{5}}{2} = e^{x}$
$x = ln \frac{3 + \sqrt{5}}{2}, ln \frac{3 - \sqrt{5}}{2}$

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