The number of real solutions of the equation | vedclass

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Question

(D) Given equation: $e^{4x} + 4e^{3x} - 58e^{2x} + 4e^{x} + 1 = 0$.Divide by $e^{2x}$ (since $e^{2x} 
eq 0$):$e^{2x} + 4e^{x} - 58 + 4e^{-x} + e^{-2x

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Given equation: $e^{4x} + 4e^{3x} - 58e^{2x} + 4e^{x} + 1 = 0$.Divide by $e^{2x}$ (since $e^{2x} 
eq 0$):$e^{2x} + 4e^{x} - 58 + 4e^{-x} + e^{-2x} = 0$.Rearrange terms:$(e^{2x} + e^{-2x}) + 4(e^{x} + e^{-x}) - 58 = 0$.Let $t = e^{x} + e^{-x}$. Note that for real $x$,$t \geq 2$.Since $(e^{x} + e^{-x})^{2} = e^{2x} + e^{-2x} + 2$,we have $e^{2x} + e^{-2x} = t^{2} - 2$.Substituting into the equation:$(t^{2} - 2) + 4t - 58 = 0 \implies t^{2} + 4t - 60 = 0$.Factor the quadratic:$(t + 10)(t - 6) = 0$.So,$t = -10$ or $t = 6$.Since $t = e^{x} + e^{-x} \geq 2$,we discard $t = -10$.Thus,$e^{x} + e^{-x} = 6$.$e^{2x} - 6e^{x} + 1 = 0$.Let $u = e^{x}$. Then $u^{2} - 6u + 1 = 0$.The discriminant $D = (-6)^{2} - 4(1)(1) = 36 - 4 = 32 > 0$.Since the product of roots is $1 > 0$ and the sum of roots is $6 > 0$,both roots $u_{1}, u_{2}$ are positive.Since $u = e^{x} > 0$ for any real $x$,both roots provide valid real solutions for $x = \ln(u)$.Therefore,there are $2$ real solutions.

The number of real solutions of the equation $e^{4x} + 4e^{3x} - 58e^{2x} + 4e^{x} + 1 = 0$ is:

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