The number of real solutions of the equation e ^ 4x + 4 e ^ 3x - 58...

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

Let $f\left(x\right) = {e}^{2x}\left({e}^{2x} + \dfrac{1}{{e}^{2x}} + 4\left({e}^{x} + \dfrac{1}{{e}^{x}}\right) - 58\right)$

Now let ${e}^{x} + \dfrac{1}{{e}^{x}} = t$

i.e. ${t}^{2} + 4t - 58 = 0$

$t = \dfrac{- 4 \pm \sqrt{16 + 4 \times 58}}{2}$

$= \dfrac{- 4 \pm 2\sqrt{62}}{2}$

$t = - 2 + 2\sqrt{62}$ is possible and $t = - 2 - 2\sqrt{62}$ is not possible as $t \ge 2$

i.e. ${e}^{x} + \dfrac{1}{{e}^{x}} = - 2 + 2\sqrt{62}$

${e}^{2x} - \left(- 2 + 2\sqrt{62}\right){e}^{x} + 1 = 0$

For real solution, discriminant $248 - 8\sqrt{62} > 0$

Also $\dfrac{- b}{2a} = \dfrac{- 2 + 2\sqrt{62}}{2} = \sqrt{62} - 1 > 0$

Hence, both roots of the equation are positive real roots.