Question

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

Answer 1

Answer: 2

$e^{4x}+4e^{3x}-58e^{2x}+4e^x+1=0$
Let $f(x)=e^{2x}\left(e^{2x}+\frac{1}{e^{2x}}+4\left(e^x+\frac{1}{e^x}\right)-58\right)$
$e^x+\frac{1}{e^x}$
Let $h(t)=t^2+4t-58=0$
$t=\frac{-4\pm \sqrt{16+4.58}}{2}$
$\frac{-4\pm 2\sqrt{62}}{2}$
$t_1=-2+2\sqrt{62}$
$t_2=-2-2\sqrt{62}$ (not possible)
$t\ge 2$
$e^x+\frac{1}{e^x}=-2+2\sqrt{62}$
$e^{2x}-(-2+2\sqrt{62})e^x+1=0$
$(-2+2\sqrt{62})-4$
$4+4.62-8\sqrt{62}-4$
$248-8\sqrt{62}>0$
$\frac{-b}{2a}>0$
both roots are positive
$2$ real roots