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

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

1. Perform Algebraic Substitution

Let $$y = e^x$$. Since $$e^x > 0$$ for all real values of $$x$$, we need to find the positive real roots where $$y > 0$$ for the polynomial:

$$y^4 - y^3 - 4y^2 - y + 1 = 0$$

2. Simplify Using Reciprocal Symmetry

Since $$y = 0$$ is not a solution, divide the entire equation by $$y^2$$:

$$\left(y^2 + \frac{1}{y^2}\right) - \left(y + \frac{1}{y}\right) - 4 = 0$$

Let $$t = y + \frac{1}{y}$$. Squaring both sides gives $$t^2 = y^2 + \frac{1}{y^2} + 2$$, which means $$y^2 + \frac{1}{y^2} = t^2 - 2$$. Substituting these terms into the equation yields:

$$(t^2 - 2) - t - 4 = 0$$

$$t^2 - t - 6 = 0$$

3. Solve for t

Factor the quadratic equation:

$$(t - 3)(t + 2) = 0$$

This gives two possible values: $$t = 3$$ or $$t = -2$$.

4. Analyze Constraints and Solve for x

By the Arithmetic Mean-Geometric Mean inequality, for any positive real number $$y$$:

$$t = y + \frac{1}{y} \geq 2$$

Case 1: $$t = -2$$

This case is rejected because $$t$$ must be greater than or equal to 2.

Case 2: $$t = 3$$

This is valid since $$3 \geq 2$$.

$$y + \frac{1}{y} = 3 \implies y^2 - 3y + 1 = 0$$

Using the quadratic formula to solve for $$y$$:

$$y = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(1)}}{2} = \frac{3 \pm \sqrt{5}}{2}$$

Both values are strictly positive. Since $$x = \ln(y)$$, each positive value of $$y$$ yields exactly one unique real value for $$x$$.

Final Answer

The equation has exactly 2 real roots.