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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.

Learn how to solve the exponential equation 2^x - 5 * 2^{-x} + 4 = 0 step-by-step. This problem covers key concepts in exponential and quadratic equations, including substitution and the quadratic formula. Suitable for students in Grades 10-12.

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