Key Concepts and Formulas

• Reciprocal Equations: Equations where the coefficients are symmetric. A common strategy is to divide by a power of the variable and make a substitution to simplify.
• Exponential Functions: The function $e^x$ is always positive for real $x$. Therefore, if $e^x = t$, then $t > 0$.
• Quadratic Formula: The solutions to the quadratic equation $ax^2 + bx + c = 0$ are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step-by-Step Solution

Step 1: Initial Substitution to Convert to a Polynomial

We are given the equation $e^{4x} + 8e^{3x} + 13e^{2x} - 8e^x + 1 = 0$. To simplify, we make the substitution $t = e^x$. Since $x \in \mathbb{R}$, we know that $t > 0$. Substituting, we get:

$t^4 + 8t^3 + 13t^2 - 8t + 1 = 0$

Step 2: Dividing by $t^2$ to Exploit Reciprocal Nature

This is a reciprocal equation because the coefficients are symmetric. Since $t > 0$, we can divide the equation by $t^2$ without losing any solutions. This allows us to group terms effectively:

$\frac{t^4}{t^2} + \frac{8t^3}{t^2} + \frac{13t^2}{t^2} - \frac{8t}{t^2} + \frac{1}{t^2} = 0$

$t^2 + 8t + 13 - \frac{8}{t} + \frac{1}{t^2} = 0$

Step 3: Grouping Terms for Further Simplification

We rearrange and group terms to prepare for another substitution:

$\left(t^2 + \frac{1}{t^2}\right) + 8\left(t - \frac{1}{t}\right) + 13 = 0$

Step 4: Second Substitution to Obtain a Quadratic

Let $z = t - \frac{1}{t}$. Then, $z^2 = \left(t - \frac{1}{t}\right)^2 = t^2 - 2 + \frac{1}{t^2}$, so $t^2 + \frac{1}{t^2} = z^2 + 2$. Substituting these into our equation, we get:

$(z^2 + 2) + 8z + 13 = 0$

$z^2 + 8z + 15 = 0$

Step 5: Solving the Quadratic for z

We factor the quadratic equation in $z$:

$(z + 3)(z + 5) = 0$

This gives us two possible values for $z$:

$z = -3 \quad \text{or} \quad z = -5$

Step 6: Back-Substituting to Find t

We substitute back $z = t - \frac{1}{t}$ to find the values of $t$.

• Case 1: $z = -3$

$t - \frac{1}{t} = -3$

Multiplying by $t$, we get:

$t^2 - 1 = -3t$

$t^2 + 3t - 1 = 0$

• Case 2: $z = -5$

$t - \frac{1}{t} = -5$

Multiplying by $t$, we get:

$t^2 - 1 = -5t$

$t^2 + 5t - 1 = 0$

Step 7: Solving for t and Applying the Constraint t > 0

We use the quadratic formula to solve for $t$ in each case. Remember that $t = e^x > 0$, so we discard any negative solutions.

• For $t^2 + 3t - 1 = 0$:

$t = \frac{-3 \pm \sqrt{3^2 - 4(1)(-1)}}{2(1)} = \frac{-3 \pm \sqrt{13}}{2}$

Since $t > 0$, we take the positive root:

$t_1 = \frac{-3 + \sqrt{13}}{2} \approx \frac{-3 + 3.6}{2} = 0.3 > 0$

• For $t^2 + 5t - 1 = 0$:

$t = \frac{-5 \pm \sqrt{5^2 - 4(1)(-1)}}{2(1)} = \frac{-5 \pm \sqrt{29}}{2}$

Since $t > 0$, we take the positive root:

$t_2 = \frac{-5 + \sqrt{29}}{2} \approx \frac{-5 + 5.4}{2} = 0.2 > 0$

We have two valid positive values for $t$.

Step 8: Finding the Solutions for x

Since $t = e^x$, we have $x = \ln(t)$. Therefore, the two solutions for $x$ are:

$x_1 = \ln\left(\frac{-3 + \sqrt{13}}{2}\right)$

$x_2 = \ln\left(\frac{-5 + \sqrt{29}}{2}\right)$

Step 9: Analyzing the Nature of the Solutions

Since $0 < \frac{-3 + \sqrt{13}}{2} < 1$ and $0 < \frac{-5 + \sqrt{29}}{2} < 1$, both $x_1$ and $x_2$ are negative (since $\ln(t) < 0$ if $0 < t < 1$).

Step 10: Summary of Solutions

We have found two real solutions for $x$, and both solutions are negative.

Common Mistakes & Tips

• Remember the constraint $t > 0$ when solving for $t$. Discard any negative solutions.
• Be careful with algebraic manipulations, especially when squaring or taking square roots.
• Recognize the reciprocal structure of the equation to apply the appropriate solution technique.

Summary

By making appropriate substitutions and solving the resulting quadratic equations, we found two real solutions for $x$. Since both corresponding $t$ values were between 0 and 1, we concluded that both solutions for $x$ are negative. This corresponds to option (A).

Final Answer

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