The problem is asking for the sum of the roots of the exponential equation:

$2^x - 5 \cdot 2^{-x} + 4 = 0$

Let's solve this step by step.

Step 1: Simplify the Equation

First, we can make a substitution to simplify the equation. Let:

$y = 2^x$

Then, $2^{-x}$ can be rewritten as:

$2^{-x} = \frac{1}{2^x} = \frac{1}{y}$

Substituting into the equation:

$y - 5 \cdot \frac{1}{y} + 4 = 0$

Multiply the entire equation by $y$ to eliminate the fraction:

$y^2 + 4y - 5 = 0$

Step 2: Solve the Quadratic Equation

Now we solve the quadratic equation:

$y^2 + 4y - 5 = 0$

The solutions to this quadratic equation can be found using the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1$, $b = 4$, and $c = -5$. Plugging in these values:

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

$y = \frac{-4 \pm \sqrt{16 + 20}}{2}$

$y = \frac{-4 \pm \sqrt{36}}{2}$

$y = \frac{-4 \pm 6}{2}$

So, the two solutions for $y$ are:

$y = \frac{2}{2} = 1 \quad \text{or} \quad y = \frac{-10}{2} = -5$

However, $y = 2^x$ and since $2^x$ must be positive, we discard $y = -5$. Therefore, $y = 1$.

Step 3: Find $x$

If $y = 1$, then:

$2^x = 1$

This implies:

$x = 0$

Thus, the only root is $x = 0$, and the sum of the roots is simply $0$.

Answer

The correct answer is:

$\boxed{0}$

Would you like more details or have any questions about this solution?

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Here are five related questions that expand on the topic:

1. How do you solve a quadratic equation using the quadratic formula?
2. What is the importance of discarding non-valid solutions in exponential equations?
3. Can you derive the logarithmic function from exponential equations?
4. How do we handle negative solutions in real-world exponential problems?
5. How does substitution help simplify complex algebraic equations?

Tip: When solving exponential equations, always check the feasibility of the solutions, especially when dealing with expressions like $2^x$, which must be positive.