Examine whether the following numbers are rational or irrational :
`(2sqrt3 + 3sqrt2) ^(2)`

Answer

To determine whether the expression \((2\sqrt{3} + 3\sqrt{2})^2\) is rational or irrational, we will follow these steps:

### Step 1: Expand the Expression
We start by expanding the expression using the formula \((a + b)^2 = a^2 + 2ab + b^2\).

Let:
- \(a = 2\sqrt{3}\)
- \(b = 3\sqrt{2}\)

Now, we can expand:
\[ (2\sqrt{3} + 3\sqrt{2})^2 = (2\sqrt{3})^2 + 2(2\sqrt{3})(3\sqrt{2}) + (3\sqrt{2})^2 \]

### Step 2: Calculate Each Term
Now, we calculate each term in the expansion:

1. \((2\sqrt{3})^2 = 4 \cdot 3 = 12\)
2. \(2(2\sqrt{3})(3\sqrt{2}) = 2 \cdot 2 \cdot 3 \cdot \sqrt{3} \cdot \sqrt{2} = 12\sqrt{6}\)
3. \((3\sqrt{2})^2 = 9 \cdot 2 = 18\)

### Step 3: Combine the Results
Now we combine all the results from the previous step:
\[ (2\sqrt{3} + 3\sqrt{2})^2 = 12 + 12\sqrt{6} + 18 \]
\[ = 30 + 12\sqrt{6} \]

### Step 4: Determine Rationality
Now we need to examine whether \(30 + 12\sqrt{6}\) is rational or irrational.

- The number \(30\) is a rational number.
- The term \(12\sqrt{6}\) involves \(\sqrt{6}\), which is known to be an irrational number because \(6\) is not a perfect square.

Since the sum of a rational number and an irrational number is always irrational, we conclude that:
\[ 30 + 12\sqrt{6} \text{ is irrational.} \]

### Final Conclusion
Thus, the expression \((2\sqrt{3} + 3\sqrt{2})^2\) is an **irrational number**.

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