The equation $\mathrm{e}^{4 x}+8 \mathrm{e}^{3 x}+13 \mathrm{e}^{2 x}-8 \mathrm{e}^{x}+1=0, x \in \mathbb{R}$ has :

$e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^{x}+1=0$ <br/><br/>Let $\mathrm{e}^{\mathrm{x}}=\mathrm{t}$ <br/><br/>Now, $\mathrm{t}^{4}+8 \mathrm{t}^{3}+13 \mathrm{t}^{2}-8 \mathrm{t}+1=0$ <br/><br/>Dividing equation by $\mathrm{t}^{2}$ <br/><br/>$$ \begin{aligned} & t^{2}+8 t+13-\frac{8}{t}+\frac{1}{t^{2}}=0 \\\\ & t^{2}+\frac{1}{t^{2}}+8\left(t-\frac{1}{t}\right)+13=0 \\\\ & \left(t-\frac{1}{t}\right)^{2}+2+8\left(t-\frac{1}{t}\right)+13=0 \end{aligned} $$ <br/><br/>Let $\mathrm{t}-\frac{1}{\mathrm{t}}=\mathrm{z}$ <br/><br/>$z^{2}+8 z+15=0$ <br/><br/>$$ \begin{aligned} & (z+3)(z+5)=0 \end{aligned} $$ <br/><br/>$z=-3 \text { or } z=-5$ <br/><br/>So, $\mathrm{t}-\frac{1}{\mathrm{t}}=-3$ or $\mathrm{t}-\frac{1}{\mathrm{t}}=-5$ <br/><br/>$t^{2}+3 t-1=0$ or $t^{2}+5 t-1=0$ <br/><br/>$\mathrm{t}=\frac{-3 \pm \sqrt{13}}{2}$ or $\mathrm{t}=\frac{-5 \pm \sqrt{29}}{2}$ <br/><br/>as $t=e^{x}$ so $t$ must be positive, <br/><br/>$t=\frac{\sqrt{13}-3}{2} \text { or } \frac{\sqrt{29}-5}{2}$ <br/><br/>So, $x=\ln \left(\frac{\sqrt{13}-3}{2}\right)$ or $x=\ln \left(\frac{\sqrt{29}-5}{2}\right)$ <br/><br/>Hence two solutions and both are negative.