The equation e^4 x+8 e^3 x+13 e^2 x-8 e^x+1=0, x R has :

Question

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 :

Accepted Answer

$e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^{x}+1=0$

Let $\mathrm{e}^{\mathrm{x}}=\mathrm{t}$

Now, $\mathrm{t}^{4}+8 \mathrm{t}^{3}+13 \mathrm{t}^{2}-8 \mathrm{t}+1=0$

Dividing equation by $\mathrm{t}^{2}$

$$
\begin{aligned}
& t^{2}+8 t+13-\frac{8}{t}+\frac{1}{t^{2}}=0 \\
& t^{2}+\frac{1}{t^{2}}+8\left(t-\frac{1}{t}\right)+13=0 \\
& \left(t-\frac{1}{t}\right)^{2}+2+8\left(t-\frac{1}{t}\right)+13=0
\end{aligned}
$$

Let $\mathrm{t}-\frac{1}{\mathrm{t}}=\mathrm{z}$

$z^{2}+8 z+15=0$

$$
\begin{aligned}
& (z+3)(z+5)=0
\end{aligned}
$$

$$
z=-3 \text { or } z=-5
$$

So, $\mathrm{t}-\frac{1}{\mathrm{t}}=-3$ or $\mathrm{t}-\frac{1}{\mathrm{t}}=-5$

$t^{2}+3 t-1=0$ or $t^{2}+5 t-1=0$

$\mathrm{t}=\frac{-3 \pm \sqrt{13}}{2}$ or $\mathrm{t}=\frac{-5 \pm \sqrt{29}}{2}$

as $t=e^{x}$ so $t$ must be positive,

$$
t=\frac{\sqrt{13}-3}{2} \text { or } \frac{\sqrt{29}-5}{2}
$$

So, $x=\ln \left(\frac{\sqrt{13}-3}{2}\right)$ or $x=\ln \left(\frac{\sqrt{29}-5}{2}\right)$

Hence two solutions and both are negative.

Solutions · 1