The number of real roots of the equation e^4x + 2e^3x - e^x - 6 = 0 is :

Question

The number of real roots of the equation ${e^{4x}} + 2{e^{3x}} - {e^x} - 6 = 0$ is :

Accepted Answer

Let ${e^x} = t > 0$

$f(t) = {t^4} + 2{t^3} - t - 6 = 0$

$f'(t) = 4{t^3} + 6{t^2} - 1$

$f''(t) = 12{t^2} + 12t > 0$

$f(0) =  - 6,f(1) =  - 4,f(2) = 24$

$$\Rightarrow$$ Number of real roots = 1

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