The number of real roots of the equation math | vedclass

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Question

$t^{4}-t^{3}-4 t^{2}-t+1=0, e^{x}=t>0$

$\Rightarrow t^{2}-t-4-\frac{1}{t}+\frac{1}{t^{2}}=0$

$\Rightarrow \alpha^{2}-\alpha-6=0, \alpha=t+\fr

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$t^{4}-t^{3}-4 t^{2}-t+1=0, e^{x}=t>0$

$\Rightarrow t^{2}-t-4-\frac{1}{t}+\frac{1}{t^{2}}=0$

$\Rightarrow \alpha^{2}-\alpha-6=0, \alpha=t+\frac{1}{t} \geq 2$

$\Rightarrow \alpha=3,-2(	ext { reject })$

$\Rightarrow t+\frac{1}{t}=3$

$\Rightarrow$ The number of real roots $=2$

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 $.....$

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