Evaluate : `(i)sqrt(3-2sqrt(2))" "(ii)sqrt(9+6sqrt(2))`

Explanation

To solve the problem, we need to evaluate the two expressions given:

(i) \(\sqrt{3 - 2\sqrt{2}}\)
(ii) \(\sqrt{9 + 6\sqrt{2}}\)

Let's break down each expression step by step.

### Part (i): Evaluate \(\sqrt{3 - 2\sqrt{2}}\)

1. **Set the expression equal to \(x\)**:
\[ x = \sqrt{3 - 2\sqrt{2}} \]

2. **Rewrite \(3\) as \(2 + 1\)**:
\[ x = \sqrt{(2 + 1) - 2\sqrt{2}} \]

3. **Recognize that \(2\) can be expressed as \((\sqrt{2})^2\) and \(1\) as \((\sqrt{1})^2\)**:
\[ x = \sqrt{(\sqrt{2})^2 + (\sqrt{1})^2 - 2(\sqrt{2})(\sqrt{1})} \]

4. **Identify the expression as a perfect square**:
This matches the formula \(a^2 + b^2 - 2ab = (a - b)^2\), where \(a = \sqrt{2}\) and \(b = \sqrt{1}\):
\[ x = \sqrt{(\sqrt{2} - \sqrt{1})^2} \]

5. **Take the square root**:
\[ x = \sqrt{2} - \sqrt{1} \]

6. **Simplify**:
\[ x = \sqrt{2} - 1 \]

### Final answer for part (i):
\[ \sqrt{3 - 2\sqrt{2}} = \sqrt{2} - 1 \]

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### Part (ii): Evaluate \(\sqrt{9 + 6\sqrt{2}}\)

1. **Set the expression equal to \(y\)**:
\[ y = \sqrt{9 + 6\sqrt{2}} \]

2. **Rewrite \(9\) as \(6 + 3\)**:
\[ y = \sqrt{(6 + 3) + 6\sqrt{2}} \]

3. **Recognize that \(6\) can be expressed as \((\sqrt{6})^2\) and \(3\) as \((\sqrt{3})^2\)**:
\[ y = \sqrt{(\sqrt{6})^2 + (\sqrt{3})^2 + 2(\sqrt{6})(\sqrt{3})} \]

4. **Identify the expression as a perfect square**:
This matches the formula \(a^2 + b^2 + 2ab = (a + b)^2\), where \(a = \sqrt{6}\) and \(b = \sqrt{3}\):
\[ y = \sqrt{(\sqrt{6} + \sqrt{3})^2} \]

5. **Take the square root**:
\[ y = \sqrt{6} + \sqrt{3} \]

### Final answer for part (ii):
\[ \sqrt{9 + 6\sqrt{2}} = \sqrt{6} + \sqrt{3} \]

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### Summary of Answers:
(i) \(\sqrt{3 - 2\sqrt{2}} = \sqrt{2} - 1\)
(ii) \(\sqrt{9 + 6\sqrt{2}} = \sqrt{6} + \sqrt{3}\)

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