The number of real solutions of the equation | vedclass

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Question

$e^{4 x}+4 e^{3 x}-58 e^{2 x}+4 e^{x}+1=0$

Let $f(x)=e^{2 x}\left(e^{2 x}+\frac{1}{e^{2 x}}+4\left(e^{x}+\frac{1}{e^{x}}\right)-58\right)$

$e^{x

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$e^{4 x}+4 e^{3 x}-58 e^{2 x}+4 e^{x}+1=0$

Let $f(x)=e^{2 x}\left(e^{2 x}+\frac{1}{e^{2 x}}+4\left(e^{x}+\frac{1}{e^{x}}\right)-58\right)$

$e^{x}+\frac{1}{e^{x}}$

Let $h(t)=t^{2}+4 t-58=0$

$t =\frac{-4 \pm \sqrt{16+4.58}}{2}$

$\frac{-4 \pm 2 \sqrt{62}}{2}$

$t _{1}=-2+2 \sqrt{62}$

$t _{2}=-2-2 \sqrt{62}$ (not possible)

$t \geq 2$

$e ^{ x }+\frac{1}{ e ^{ x }}=-2+2 \sqrt{62}$

$e ^{2 x }-(-2+2 \sqrt{62}) e ^{ x }+1=0$

$(-2+2 \sqrt{62})-4$

$4+4.62-8 \sqrt{62}-4$

$248-8 \sqrt{62}>0$

$\frac{- b }{2 a }>0$

both roots are positive

$2$ real roots

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

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