The number of real roots of the equation e^{4 | vedclass

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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Given equation: $e^{4x} + e^{3x} - 4e^{2x} + e^x + 1 = 0$.Divide the entire equation by $e^{2x}$ (since $e^{2x} 
eq 0$ for all real $x$):$e^{2x} + e^x - 4 + e^{-x} + e^{-2x} = 0$.Rearrange the terms:$(e^{2x} + e^{-2x}) + (e^x + e^{-x}) - 4 = 0$.Let $u = e^x + e^{-x}$. Note that $u^2 = (e^x + e^{-x})^2 = e^{2x} + 2 + e^{-2x}$,so $e^{2x} + e^{-2x} = u^2 - 2$.Substitute this into the equation:$(u^2 - 2) + u - 4 = 0 \Rightarrow u^2 + u - 6 = 0$.Factor the quadratic equation:$(u + 3)(u - 2) = 0$.This gives $u = -3$ or $u = 2$.Case $1$: $e^x + e^{-x} = -3$. Since $e^x > 0$ and $e^{-x} > 0$,by $AM$-$GM$ inequality,$e^x + e^{-x} \geq 2$. Thus,$u = -3$ has no real solutions.Case $2$: $e^x + e^{-x} = 2$. By $AM$-$GM$ inequality,$e^x + e^{-x} = 2$ only when $e^x = e^{-x}$,which implies $e^{2x} = 1$,so $x = 0$.Thus,there is only $1$ real root.

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

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