The number of real roots of the equation e^{4 | vedclass

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Question

(B) Given equation: $e^{4x} - e^{3x} - 4e^{2x} - e^{x} + 1 = 0$. Divide by $e^{2x}$ (since $e^{2x} > 0$): $e^{2x} - e^{x} - 4 - e^{-x} + e^{-2x} = 0$.

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given equation: $e^{4x} - e^{3x} - 4e^{2x} - e^{x} + 1 = 0$. Divide by $e^{2x}$ (since $e^{2x} > 0$): $e^{2x} - e^{x} - 4 - e^{-x} + e^{-2x} = 0$. Group terms: $(e^{2x} + e^{-2x}) - (e^{x} + e^{-x}) - 4 = 0$. Let $t = e^{x} > 0$. Then $e^{x} + e^{-x} = t + \frac{1}{t} = \alpha$,where $\alpha \geq 2$. Note that $e^{2x} + e^{-2x} = (t + \frac{1}{t})^2 - 2 = \alpha^2 - 2$. Substituting into the equation: $(\alpha^2 - 2) - \alpha - 4 = 0$. $\alpha^2 - \alpha - 6 = 0$. $(\alpha - 3)(\alpha + 2) = 0$. Since $\alpha \geq 2$,we have $\alpha = 3$. Now,$t + \frac{1}{t} = 3 \Rightarrow t^2 - 3t + 1 = 0$. The discriminant $D = (-3)^2 - 4(1)(1) = 5 > 0$. Since $t = e^{x} > 0$ and the product of roots is $1$ and sum is $3$,both roots for $t$ are positive. Thus,there are $2$ real roots for $x$.

The number of real roots of the equation $e^{4x} - e^{3x} - 4e^{2x} - e^{x} + 1 = 0$ is equal to $.....$

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