State, whether the following numbers are rational or not:
`(2 + sqrt(2))^(2)`

Answer

To determine whether the number \((2 + \sqrt{2})^2\) is rational or irrational, we can follow these steps:

### Step 1: Expand the expression using the formula for the square of a binomial.
We use the formula \((a + b)^2 = a^2 + 2ab + b^2\), where \(a = 2\) and \(b = \sqrt{2}\).

\[ (2 + \sqrt{2})^2 = 2^2 + 2 \cdot 2 \cdot \sqrt{2} + (\sqrt{2})^2 \]

### Step 2: Calculate each term.
- Calculate \(2^2\):
\[ 2^2 = 4 \]

- Calculate \(2 \cdot 2 \cdot \sqrt{2}\):
\[ 2 \cdot 2 \cdot \sqrt{2} = 4\sqrt{2} \]

- Calculate \((\sqrt{2})^2\):
\[ (\sqrt{2})^2 = 2 \]

### Step 3: Combine the results.
Now, we add the results from Step 2 together:
\[ (2 + \sqrt{2})^2 = 4 + 4\sqrt{2} + 2 \]
\[ = 6 + 4\sqrt{2} \]

### Step 4: Analyze the expression \(6 + 4\sqrt{2}\).
- The number \(6\) is rational.
- The term \(4\sqrt{2}\) is irrational because \(\sqrt{2}\) is an irrational number.

### Step 5: Determine the nature of the sum.
According to the properties of rational and irrational numbers, the sum of a rational number and an irrational number is always irrational. Therefore, we conclude that:

\[ (2 + \sqrt{2})^2 = 6 + 4\sqrt{2} \text{ is irrational.} \]

### Final Conclusion:
Thus, \((2 + \sqrt{2})^2\) is an **irrational number**.

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