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

Option: 1

four solutions two of which are negative
 


Option: 2

two solutions and only one of them is negative
 


Option: 3

two solutions and both are negative
 


Option: 4

no solution


Answers (1)

best_answer

$$ \begin{aligned} & \mathrm{e}^{4 x}+8 \mathrm{e}^{3 x}+13 \mathrm{e}^{2 x}+13 \mathrm{e}^{2 x}-8 \mathrm{e}^{\mathrm{x}}+1=0, x \in R \\ \end{aligned}

$$ \begin{aligned} & \text { Let } \mathrm{e}^{\mathrm{x}}=\mathrm{t}>0 \& \mathrm{x}=\operatorname{lnt} \\ \end{aligned}

$$ \begin{aligned} & \mathrm{t}^4+8 t^3+13 t^2-8 t+1=0 \end{aligned}
Dividing 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 \end{aligned}
Let \mathrm{t}-\frac{1}{\mathrm{t}}=\mathrm{u} \Rightarrow \mathrm{t}^2+\frac{1}{t^2}-2 \mathrm{u}^2
\begin{array}{ll} \Rightarrow \mathrm{t}^2+\frac{1}{\mathrm{t}^2}=\mathrm{u}^2+2 & \\ \end{array}

\begin{array}{ll} \mathrm{u}^2+2+8 \mathrm{u}+13=0 & \\ (\mathrm{u}+3)(\mathrm{u}+5)=0 & \\ \mathrm{u}=-3 \& \mathrm{u}=-5 & \\ \mathrm{t}-\frac{1}{\mathrm{t}}=-3 & \mathrm{t}-\frac{1}{\mathrm{t}}=-5 \\ \mathrm{t}^2+3 \mathrm{t}-1=0 & \mathrm{t}^2+5 \mathrm{t}-1=0 \end{array}

\mathrm{0< \alpha _{1}< 1}                                                  \mathrm{0< \alpha _{2}< 1}

\mathrm{\Rightarrow x_1=\ln \alpha_1<0 \quad \Rightarrow \mathrm{x}_2=\ln \alpha_2<0}

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