The sum of the roots of the equation e^{4t} - | vedclass

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Question

(B) Given the equation: $e^{4t} - 10e^{3t} + 29e^{2t} - 20e^t + 4 = 0$. Let $x = e^t$. Then the equation becomes $x^4 - 10x^3 + 29x^2 - 20x + 4 = 0$.

Vedclass · Accepted Answer

$2\log_e 2$

Vedclass · Answer

(B) Given the equation: $e^{4t} - 10e^{3t} + 29e^{2t} - 20e^t + 4 = 0$. Let $x = e^t$. Then the equation becomes $x^4 - 10x^3 + 29x^2 - 20x + 4 = 0$. Let the roots of the equation in $t$ be $t_1, t_2, t_3, t_4$. Then the roots of the equation in $x$ are $x_1 = e^{t_1}, x_2 = e^{t_2}, x_3 = e^{t_3}, x_4 = e^{t_4}$. By Vieta's formulas,the product of the roots of the polynomial $x^4 - 10x^3 + 29x^2 - 20x + 4 = 0$ is given by the constant term: $x_1 \cdot x_2 \cdot x_3 \cdot x_4 = 4$. Substituting back $e^t$: $e^{t_1} \cdot e^{t_2} \cdot e^{t_3} \cdot e^{t_4} = 4$. $e^{(t_1 + t_2 + t_3 + t_4)} = 4$. Taking the natural logarithm on both sides: $t_1 + t_2 + t_3 + t_4 = \log_e 4$. Since $4 = 2^2$,we have: $t_1 + t_2 + t_3 + t_4 = \log_e 2^2 = 2\log_e 2$.

The sum of the roots of the equation $e^{4t} - 10e^{3t} + 29e^{2t} - 20e^t + 4 = 0$ is

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