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Let's solve the problem step by step.

### 1. **Solve the Equation \( 4^x + \frac{5}{2} - \frac{9}{2^{2x}} = 0 \)**

Given that \( \frac{1}{a} \) is a solution of this equation, let's solve for \( x \).

Rewriting the equation:

\[
4^x + \frac{5}{2} - \frac{9}{2^{2x}} = 0
\]

Notice that \( 4^x \) can be written as \( (2^2)^x = (2^x)^2 \), and the term \( \frac{9}{2^{2x}} \) can be written as \( \frac{9}{(2^x)^2} \).

Let's substitute \( y = 2^x \), then the equation becomes:

\[
y^2 + \frac{5}{2} - \frac{9}{y^2} = 0
\]

Multiply the entire equation by \( 2y^2 \) to eliminate the fractions:

\[
2y^4 + 5y^2 - 18 = 0
\]

This is a quadratic equation in \( y^2 \). Let \( z = y^2 \), then:

\[
2z^2 + 5z - 18 = 0
\]

Solve this quadratic equation using the quadratic formula:

\[
z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where \( a = 2 \), \( b = 5 \), and \( c = -18 \):

\[
z = \frac{-5 \pm \sqrt{5^2 - 4(2)(-18)}}{2(2)}
\]
\[
z = \frac{-5 \pm \sqrt{25 + 144}}{4}
\]
\[
z = \frac{-5 \pm \sqrt{169}}{4}
\]
\[
z = \frac{-5 \pm 13}{4}
\]

This gives us two possible solutions for \( z \):

\[
z = \frac{8}{4} = 2 \quad \text{or} \quad z = \frac{-18}{4} = -4.5
\]

Since \( z = y^2 = 2^x \), and \( y^2 \) must be positive, we discard \( z = -4.5 \). So:

\[
y^2 = 2 \quad \text{which means} \quad 2^x = \sqrt{2}
\]

Taking the logarithm base 2:

\[
x = \frac{1}{2}
\]

Thus, \( x = \frac{1}{2} \). 

Given that \( \frac{1}{a} = \frac{1}{2} \), it implies \( a = 2 \).

### 2. **Calculate the Equivalent \( K = \sqrt{2^a + 2} \)**

Substitute \( a = 2 \) into the equation:

\[
K = \sqrt{2^2 + 2} = \sqrt{4 + 2} = \sqrt{6}
\]

So, the correct answer is:

**Option C) 6**

### Questions or Clarifications?
Would you like any further details or explanations?

### Related Questions
1. What are the steps for solving a quadratic equation?
2. How do you simplify expressions involving square roots?
3. What is the significance of the quadratic formula in solving equations?
4. How does logarithm help in solving exponential equations?
5. Can a negative number ever be a solution for a square in a quadratic equation?
6. How do you rewrite exponential expressions in terms of base 2?
7. What are the rules for solving logarithmic equations?
8. How do you verify the solution of a quadratic equation?

### Tip
Always double-check the steps when applying the quadratic formula to avoid mistakes in calculations, especially when dealing with signs in the discriminant.

Learn how to solve complex exponential and quadratic equations step-by-step. This problem involves solving equations with exponential terms and using the quadratic formula. Suitable for students in Grades 10-12.

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