The equation e^{4x} + 8e^{3x} + 13e^{2x} - 8e | vedclass

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(A) Given equation: $e^{4x} + 8e^{3x} + 13e^{2x} - 8e^x + 1 = 0$Let $e^x = t$. Since $x \in R$,$t > 0$.The equation becomes $t^4 + 8t^3 + 13t^2 - 8t +

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two solutions and both are negative

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(A) Given equation: $e^{4x} + 8e^{3x} + 13e^{2x} - 8e^x + 1 = 0$Let $e^x = t$. Since $x \in R$,$t > 0$.The equation becomes $t^4 + 8t^3 + 13t^2 - 8t + 1 = 0$.Dividing by $t^2$ (as $t 
eq 0$):$t^2 + 8t + 13 - \frac{8}{t} + \frac{1}{t^2} = 0$Rearranging terms:$(t^2 + \frac{1}{t^2}) + 8(t - \frac{1}{t}) + 13 = 0$Let $z = t - \frac{1}{t}$. Then $z^2 = t^2 + \frac{1}{t^2} - 2$,so $t^2 + \frac{1}{t^2} = z^2 + 2$.Substituting into the equation:$(z^2 + 2) + 8z + 13 = 0$$z^2 + 8z + 15 = 0$$(z + 3)(z + 5) = 0$So,$z = -3$ or $z = -5$.Case $1$: $t - \frac{1}{t} = -3 \implies t^2 + 3t - 1 = 0$. Roots are $t = \frac{-3 \pm \sqrt{9 + 4}}{2} = \frac{-3 \pm \sqrt{13}}{2}$. Since $t > 0$,$t = \frac{\sqrt{13} - 3}{2}$.Case $2$: $t - \frac{1}{t} = -5 \implies t^2 + 5t - 1 = 0$. Roots are $t = \frac{-5 \pm \sqrt{25 + 4}}{2} = \frac{-5 \pm \sqrt{29}}{2}$. Since $t > 0$,$t = \frac{\sqrt{29} - 5}{2}$.Since $\sqrt{13} \approx 3.6$,$\frac{\sqrt{13}-3}{2} \approx 0.3 Since $\sqrt{29} \approx 5.38$,$\frac{\sqrt{29}-5}{2} \approx 0.19 Thus,there are two solutions and both are negative.

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

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