Find the square root of
`(i) (2+sqrt(3))/(2)`, `(ii) 12-sqrt(68+48sqrt((2)))`.

Answer

To find the square roots of the given expressions, we will solve each part step by step.

### Part (i): Find the square root of \((2 + \sqrt{3})/2\)

1. **Rewrite the Expression**:
\[ \sqrt{\frac{2 + \sqrt{3}}{2}} = \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2}} \]

2. **Simplify the Numerator**:
We need to express \(2 + \sqrt{3}\) in a form that allows us to take the square root easily. We can try to express it as a square of a binomial:
\[ 2 + \sqrt{3} = \left(\sqrt{a} + \sqrt{b}\right)^2 \]
Expanding this gives:
\[ a + b + 2\sqrt{ab} = 2 + \sqrt{3} \]
By comparing coefficients, we can set:
- \(a + b = 2\)
- \(2\sqrt{ab} = \sqrt{3}\)

From \(2\sqrt{ab} = \sqrt{3}\), we find:
\[ \sqrt{ab} = \frac{\sqrt{3}}{2} \implies ab = \frac{3}{4} \]

3. **Solve the System of Equations**:
We have the equations:
- \(a + b = 2\)
- \(ab = \frac{3}{4}\)

Let \(a\) and \(b\) be the roots of the quadratic equation:
\[ x^2 - (a+b)x + ab = 0 \implies x^2 - 2x + \frac{3}{4} = 0 \]
The discriminant is:
\[ D = 2^2 - 4 \cdot 1 \cdot \frac{3}{4} = 4 - 3 = 1 \]
The roots are:
\[ x = \frac{2 \pm 1}{2} \implies x = \frac{3}{2} \text{ or } x = \frac{1}{2} \]
Thus, \(a = \frac{3}{2}\) and \(b = \frac{1}{2}\).

4. **Substituting Back**:
Now we can write:
\[ 2 + \sqrt{3} = \left(\sqrt{\frac{3}{2}} + \sqrt{\frac{1}{2}}\right)^2 \]
Therefore:
\[ \sqrt{2 + \sqrt{3}} = \sqrt{\frac{3}{2}} + \sqrt{\frac{1}{2}} = \frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{\sqrt{3} + 1}{\sqrt{2}} \]

5. **Final Result**:
Now substituting back into our expression:
\[ \sqrt{\frac{2 + \sqrt{3}}{2}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{2}}}{\sqrt{2}} = \frac{\sqrt{3} + 1}{2} \]

### Part (ii): Find the square root of \(12 - \sqrt{68 + 48\sqrt{2}}\)

1. **Rewrite the Expression**:
First, we simplify the expression under the square root:
\[ \sqrt{68 + 48\sqrt{2}} \]

2. **Express as a Square**:
We want to express \(68 + 48\sqrt{2}\) as \((a + b\sqrt{2})^2\):
\[ (a + b\sqrt{2})^2 = a^2 + 2ab\sqrt{2} + 2b^2 \]
Comparing coefficients, we have:
- \(a^2 + 2b^2 = 68\)
- \(2ab = 48\)

From \(2ab = 48\), we find:
\[ ab = 24 \]

3. **Solve the System of Equations**:
We have:
- \(a^2 + 2b^2 = 68\)
- \(ab = 24\)

Substitute \(b = \frac{24}{a}\) into the first equation:
\[ a^2 + 2\left(\frac{24}{a}\right)^2 = 68 \]
Simplifying gives:
\[ a^2 + \frac{1152}{a^2} = 68 \]
Multiplying through by \(a^2\) leads to:
\[ a^4 - 68a^2 + 1152 = 0 \]
Let \(x = a^2\):
\[ x^2 - 68x + 1152 = 0 \]
Using the quadratic formula:
\[ x = \frac{68 \pm \sqrt{68^2 - 4 \cdot 1152}}{2} \]
Calculate the discriminant:
\[ 68^2 - 4608 = 4624 \implies \sqrt{4624} = 68 \]
Thus:
\[ x = \frac{68 \pm 68}{2} \implies x = 68 \text{ or } 0 \]
Therefore, \(a^2 = 68\) and \(b^2 = 0\).

4. **Finding \(a\) and \(b\)**:
Since \(b = 0\), we have:
\[ a = 8 \]
Thus:
\[ \sqrt{68 + 48\sqrt{2}} = 8 + 0\sqrt{2} = 8 \]

5. **Final Result**:
Now substituting back into our expression:
\[ 12 - \sqrt{68 + 48\sqrt{2}} = 12 - 8 = 4 \]
Therefore:
\[ \sqrt{4} = 2 \]

### Final Answers:
1. \(\sqrt{\frac{2 + \sqrt{3}}{2}} = \frac{\sqrt{3} + 1}{2}\)
2. \(\sqrt{12 - \sqrt{68 + 48\sqrt{2}}} = 2\)