If x is real, the number of real roots of the equation e^4x+e^3x-4e^2x+e^x+1=0 is:

Question

If $x$ is real, the number of real roots of the equation $e^{4x}+e^{3x}-4e^{2x}+e^x+1=0$ is:

Accepted Answer

Let $t=e^x>0$. Dividing by $t^2$: $t^2+\dfrac1{t^2}+t+\dfrac1t-4=0$. With $y=t+\dfrac1t$: $y^2-2+y-4=0 \implies y^2+y-6=0 \implies y=2,-3$; only $y=2$ is achievable (as $y\ge2$).

$t+\dfrac1t=2 \implies t=1 \implies x=0$ is the only solution --- 1 real root.

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