Key Concepts and Formulas

• Substitution: Replacing a complex expression with a simpler variable to simplify the equation.
• Reciprocal Equation: A polynomial equation where the coefficients are symmetric. For example, $ax^4 + bx^3 + cx^2 + bx + a = 0$.
• AM-GM Inequality: For non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. For two numbers $a, b \ge 0$, $\frac{a+b}{2} \ge \sqrt{ab}$.

Step-by-Step Solution

1. Substitution to a Polynomial Equation We are given the equation $e^{4x} - e^{3x} - 4e^{2x} - e^x + 1 = 0$. To simplify it, we substitute $t = e^x$. Since $x$ is a real number, $t$ must be a positive real number ($t > 0$). Substituting $t$ into the given equation, we get: $(e^x)^4 - (e^x)^3 - 4(e^x)^2 - e^x + 1 = 0$ $t^4 - t^3 - 4t^2 - t + 1 = 0$

2. Identifying and Solving the Reciprocal Equation The equation $t^4 - t^3 - 4t^2 - t + 1 = 0$ is a reciprocal equation of Type I, as its coefficients are symmetric ($1, -1, -4, -1, 1$).

   • Divide by $t^2$: Since $t = e^x$, we know $t > 0$, so $t \neq 0$. We can safely divide the entire equation by $t^2$ to simplify it. $\frac{t^4}{t^2} - \frac{t^3}{t^2} - \frac{4t^2}{t^2} - \frac{t}{t^2} + \frac{1}{t^2} = 0$ $t^2 - t - 4 - \frac{1}{t} + \frac{1}{t^2} = 0$
   • Grouping Terms: We group the terms as follows to prepare for another substitution: $\left(t^2 + \frac{1}{t^2}\right) - \left(t + \frac{1}{t}\right) - 4 = 0$
   • Introducing a New Variable: Let $\alpha = t + \frac{1}{t}$. This is a standard substitution for solving reciprocal equations.
     • Squaring $\alpha$, we get $t^2 + \frac{1}{t^2} + 2 = \alpha^2$, so $t^2 + \frac{1}{t^2} = \alpha^2 - 2$.
   • Condition on $\alpha$: Since $t > 0$, by the AM-GM inequality, $t + \frac{1}{t} \ge 2\sqrt{t \cdot \frac{1}{t}} = 2$, which means $\alpha \ge 2$. This is a crucial condition.
   • Substituting $\alpha$: Substitute $\alpha$ and $\alpha^2 - 2$ into the grouped equation: $(\alpha^2 - 2) - \alpha - 4 = 0$ $\alpha^2 - \alpha - 6 = 0$

3. Solving the Quadratic Equation for $\alpha$ We solve the quadratic equation $\alpha^2 - \alpha - 6 = 0$: $(\alpha - 3)(\alpha + 2) = 0$ This gives us two possible values for $\alpha$: $\alpha = 3$ or $\alpha = -2$.

4. Applying the Constraint on $\alpha$ We must apply the condition $\alpha \ge 2$.

   • $\alpha = 3$: This value satisfies $\alpha \ge 2$.
   • $\alpha = -2$: This value does not satisfy $\alpha \ge 2$, so we reject it.

5. Solving for $t$ using the valid $\alpha$ value We only consider $\alpha = 3$. Substituting back $\alpha = t + \frac{1}{t}$: $t + \frac{1}{t} = 3$

   • Multiply by $t$: To convert this into a standard quadratic equation in $t$. Since $t>0$, we can multiply without changing the solution set for positive $t$. $t^2 + 1 = 3t$ $t^2 - 3t + 1 = 0$

6. Finding the Roots for $t$ We solve the quadratic equation $t^2 - 3t + 1 = 0$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $t = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)}$ $t = \frac{3 \pm \sqrt{9 - 4}}{2}$ $t = \frac{3 \pm \sqrt{5}}{2}$ This gives two distinct real roots for $t$: $t_1 = \frac{3 + \sqrt{5}}{2}$ $t_2 = \frac{3 - \sqrt{5}}{2}$

7. Verifying that $t$ values are Positive We need to ensure that these roots for $t$ are positive, as required by $t = e^x$.

   • $t_1 = \frac{3 + \sqrt{5}}{2}$: Since $\sqrt{5}$ is positive, $3 + \sqrt{5}$ is positive, and so is $t_1 > 0$.
   • $t_2 = \frac{3 - \sqrt{5}}{2}$: Since $3 > \sqrt{5}$, $3 - \sqrt{5}$ is positive, and so is $t_2 > 0$. Both roots are positive real numbers.

8. Finding the Real Roots for $x$ For each valid positive real root of $t$, we find a unique real root for $x$ using $x = \ln(t)$.

   • For $t_1 = \frac{3 + \sqrt{5}}{2}$, we get $x_1 = \ln\left(\frac{3 + \sqrt{5}}{2}\right)$.
   • For $t_2 = \frac{3 - \sqrt{5}}{2}$, we get $x_2 = \ln\left(\frac{3 - \sqrt{5}}{2}\right)$. Since $t_1$ and $t_2$ are distinct positive real numbers, $x_1$ and $x_2$ are distinct real numbers. Therefore, there are two real roots for $x$.

Common Mistakes & Tips

• Forgetting $t > 0$: Always remember that $t = e^x$ implies $t$ must be strictly positive. Any negative or zero roots for $t$ are extraneous.
• Forgetting $\alpha \ge 2$: The condition $\alpha = t + \frac{1}{t} \ge 2$ for $t > 0$ is critical. Failing to apply it can lead to incorrectly accepting solutions.
• Incorrect Factoring The quartic $t^4 - t^3 - 4t^2 - t + 1$ does NOT factor as $(t^2 - 3t + 1)(t+1)^2$.

Summary By substituting $t=e^x$ and recognizing the equation as a reciprocal equation, we transformed it into a quadratic in $\alpha = t + \frac{1}{t}$. After solving for $\alpha$, applying the constraint $\alpha \ge 2$, and then solving for $t$, we found two distinct positive real values for $t$. Each of these positive $t$ values corresponds to a unique real root for $x$, resulting in a total of 2 real roots for the original equation.

Final Answer The final answer is \boxed{2}.