Key Concepts and Formulas

• Symmetric/Reciprocal Equations: Recognizing and manipulating equations where terms exhibit a reciprocal relationship after division by a central term.
• AM-GM Inequality: For non-negative real numbers $a_1, a_2, ..., a_n$, the Arithmetic Mean is greater than or equal to the Geometric Mean: $\frac{a_1 + a_2 + ... + a_n}{n} \ge \sqrt[n]{a_1 a_2 ... a_n}$.
• Quadratic Formula/Factoring: Methods for solving quadratic equations of the form $ax^2 + bx + c = 0$.

Step-by-Step Solution

Step 1: Rewriting the Equation

We are given the equation: $e^{4x} + e^{3x} - 4e^{2x} + e^x + 1 = 0$ We want to transform this into a form suitable for substitution. Dividing by $e^{2x}$ will reveal a symmetric structure.

• Why: Dividing by $e^{2x}$ allows us to group reciprocal terms. Also, $e^{2x} > 0$ for all real $x$, so we are not adding or removing any solutions by dividing by it.

Dividing the entire equation by $e^{2x}$: $\frac{e^{4x}}{e^{2x}} + \frac{e^{3x}}{e^{2x}} - \frac{4e^{2x}}{e^{2x}} + \frac{e^x}{e^{2x}} + \frac{1}{e^{2x}} = 0$ Simplifying the expression: $e^{2x} + e^x - 4 + e^{-x} + e^{-2x} = 0$ Rearranging the terms to group reciprocal terms: $(e^{2x} + e^{-2x}) + (e^x + e^{-x}) - 4 = 0$ Expressing negative exponents as reciprocals: $\left(e^{2x} + \frac{1}{e^{2x}}\right) + \left(e^x + \frac{1}{e^x}\right) - 4 = 0$

Step 2: Substitution and Quadratic Transformation

We will now introduce a substitution to simplify the equation further.

• Why: The terms $e^{2x} + \frac{1}{e^{2x}}$ and $e^x + \frac{1}{e^x}$ are related. Letting $z = e^x + \frac{1}{e^x}$, we can express $e^{2x} + \frac{1}{e^{2x}}$ in terms of $z$.

Let $z = e^x + \frac{1}{e^x}$. Then, $z^2 = \left(e^x + \frac{1}{e^x}\right)^2 = e^{2x} + 2(e^x)\left(\frac{1}{e^x}\right) + \frac{1}{e^{2x}} = e^{2x} + 2 + \frac{1}{e^{2x}}$ Therefore, $e^{2x} + \frac{1}{e^{2x}} = z^2 - 2$. Substituting into our equation: $(z^2 - 2) + z - 4 = 0$ Simplifying the equation: $z^2 + z - 6 = 0$

Step 3: Solving for the Intermediate Variable ($z$)

We now have a standard quadratic equation in terms of $z$.

• Why: Solving this quadratic equation will give us the possible values for our substituted expression $e^x + \frac{1}{e^x}$.

Factoring the quadratic equation $z^2 + z - 6 = 0$: We look for two numbers that multiply to -6 and add to +1. These numbers are +3 and -2. $(z + 3)(z - 2) = 0$ This yields two possible values for $z$: $z = -3 \quad \text{or} \quad z = 2$

Step 4: Analyzing the Solutions for $z$ and Finding $x$

We have two potential values for $z = e^x + \frac{1}{e^x}$. We must now determine if these values lead to real solutions for $x$.

• Understanding the Valid Range of $z$

  • Why: It is crucial to remember that $x$ must be a real number. For any real number $x$, $e^x$ is always positive. The expression $z = e^x + \frac{1}{e^x}$ is a sum of two positive numbers. We can use the AM-GM inequality to find the minimum possible value of $z$.
  • Applying AM-GM to $e^x$ and $\frac{1}{e^x}$: $\frac{e^x + \frac{1}{e^x}}{2} \ge \sqrt{e^x \cdot \frac{1}{e^x}}$ $\frac{e^x + \frac{1}{e^x}}{2} \ge \sqrt{1}$ $e^x + \frac{1}{e^x} \ge 2$
  • This means $z$ must always be greater than or equal to 2 ($z \ge 2$) for any real value of $x$.

• Case 1: $z = -3$

  • We found $z = -3$ as a solution to the quadratic equation.
  • However, as established above, $z$ must satisfy $z \ge 2$.
  • Since $-3 < 2$, this value of $z$ is not attainable for any real $x$.
  • Therefore, the equation $e^x + \frac{1}{e^x} = -3$ has no real solutions for $x$.

• Case 2: $z = 2$

  • We found $z = 2$ as a solution to the quadratic equation.
  • This value satisfies the condition $z \ge 2$, so it is a valid possibility.
  • Substitute back: $e^x + \frac{1}{e^x} = 2$.
  • To solve for $x$, we can multiply the equation by $e^x$ (which is always positive): $(e^x)^2 + 1 = 2e^x$
  • Rearrange into a quadratic form in terms of $e^x$: $(e^x)^2 - 2e^x + 1 = 0$
  • This is a perfect square trinomial: $(e^x - 1)^2 = 0$
  • Taking the square root of both sides: $e^x - 1 = 0$ $e^x = 1$
  • To find $x$, we take the natural logarithm of both sides: $x = \ln(1)$ $x = 0$
  • This gives us exactly one real solution for $x$.

Step 5: Final Count of Real Roots

By analyzing both possible values of $z$, we found that only $z=2$ leads to a real solution for $x$, which is $x=0$. Therefore, the original equation has exactly one real root.

Common Mistakes & Tips

• Tip: Always check the validity of the solutions for the intermediate variable (in this case, $z$) by considering the domain and range of the functions involved.
• Tip: The AM-GM inequality is a powerful tool for finding the minimum or maximum values of expressions, which can help determine the validity of solutions.
• Common Mistake: Forgetting that $e^x > 0$ for all real $x$. This is critical when analyzing the range of the substituted variable $z$.

Summary

The given exponential equation was transformed into a quadratic equation by dividing by $e^{2x}$ and using the substitution $z = e^x + \frac{1}{e^x}$. The key step was analyzing the possible values of $z$ in conjunction with the properties of exponential functions, particularly the AM-GM inequality, which dictates that $z \ge 2$. This analysis revealed that only one of the solutions for $z$ was valid, leading to a unique real root $x=0$ for the original equation.

Final Answer

The final answer is \boxed{1}, which corresponds to option (A).