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 _____

Answer 1

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