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

We have, $e^{4x}-e^{3x}-4e^{2x}-e^x+1=0$

Let $e^x=t$, we get

$t^4-t^3-4t^2-t+1=0$

Divide with $t^2$ on both sides

$\Rightarrow t^2-t-4-\frac{1}{t}+\frac{1}{t^2}=0$

$\Rightarrow {\left(t+\frac{1}{t}\right)}^2-\left(t+\frac{1}{t}\right)-6=0$

$$Let $\alpha =t+\frac{1}{t}\ge 2$, we get

${\alpha }^2-\alpha -6=0$

$\Rightarrow \left(\alpha -3\right)\left(\alpha +2\right)=0$

$\Rightarrow \alpha =3,-2$ (reject)

$\Rightarrow t+\frac{1}{t}=3\Rightarrow t^2-3t+1=0$

$\Rightarrow e^x=\frac{3\pm \sqrt{5}}{2}$

$\Rightarrow e^x=\frac{3\pm 2.23}{2}\left(\because \sqrt{5}\approx 2.23\right)$

$\Rightarrow e^x=\frac{5.23}{2}$ or $e^x=\frac{0.77}{2}$

$\Rightarrow e^x=2.615$ or $e^x=0.385$

$\Rightarrow$ The number of real roots $=2.$