The number of real solutions of the equation e(4x) + 4e(3x) - 58e(2x) + 4ex + 1 = 0 is

Question

The number of real solutions of the equation $e^{4 x} + 4 e^{3 x} - 58 e^{2 x} + 4 e^{x} + 1 = 0$ is

Smart Achievers · Accepted Answer

(3)

$Given equation is$

$e^{4 x} + 4 e^{3 x} - 58 e^{2 x} + 4 e^{x} + 1 = 0$

$Take, f (x) = (e^{2 x} + \frac{1}{e^{2 x}} + 4 (e^{x} + \frac{1}{e^{x}}) - 58)$

$Let e^{x} + \frac{1}{e^{x}} = p (> 0)$ $\dots (i)$

$p^{2} + 4 p - 60 = 0$

$p = 6 or p = - 10$

$Only p = 6 is allowed$

$e^{x} + \frac{1}{e^{x}} = 6$

$Two real and distinct values of x$

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