Question:
The number of real roots of the equation
$\mathrm{e}^{4 \mathrm{x}}-\mathrm{e}^{3 \mathrm{x}}-4 \mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{\mathrm{x}}+1=0$ is equal to
Solution:
$\mathrm{t}^{4}-\mathrm{t}^{3}-4 \mathrm{t}^{2}-\mathrm{t}+1=0, \mathrm{e}^{\mathrm{x}}=\mathrm{t}>0$
$\Rightarrow \mathrm{t}^{2}-\mathrm{t}-4-\frac{1}{\mathrm{t}}+\frac{1}{\mathrm{t}^{2}}=0$
$\Rightarrow \alpha^{2}-\alpha-6=0, \alpha=\mathrm{t}+\frac{1}{\mathrm{t}} \geq 2$
$\Rightarrow \alpha=3,-2$ (reject)
$\Rightarrow \mathrm{t}+\frac{1}{\mathrm{t}}=3$
$\Rightarrow$ The number of real roots $=2$
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