For the equation `2x^(2) + 6 sqrt(2)x + 1 = 0`

Options

A. roots are rational
B. if one root is `p + sqrt(q)`, then the other is `-p + sqrt(q)`
C. and if one root is `-p-sqrt(q)`, then other root `p-sqrt(q)`
D. none of these

Explanation

To solve the quadratic equation \(2x^2 + 6\sqrt{2}x + 1 = 0\), we will follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is in the standard form \(ax^2 + bx + c = 0\). Here, we have:
- \(a = 2\)
- \(b = 6\sqrt{2}\)
- \(c = 1\)

### Step 2: Calculate the discriminant
The discriminant \(D\) of a quadratic equation is given by the formula:
\[ D = b^2 - 4ac \]
Substituting the values of \(a\), \(b\), and \(c\):
\[ D = (6\sqrt{2})^2 - 4 \cdot 2 \cdot 1 \]
Calculating \(D\):
\[ D = 72 - 8 = 64 \]

### Step 3: Determine the nature of the roots
Since the discriminant \(D = 64\) is a perfect square, the roots of the equation are real and rational.

### Step 4: Calculate the roots using the quadratic formula
The roots of the quadratic equation can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{D}}{2a} \]
Substituting the values of \(b\), \(D\), and \(a\):
\[ x = \frac{-6\sqrt{2} \pm \sqrt{64}}{2 \cdot 2} \]
Calculating further:
\[ x = \frac{-6\sqrt{2} \pm 8}{4} \]
This gives us two roots:
\[ x_1 = \frac{-6\sqrt{2} + 8}{4} \quad \text{and} \quad x_2 = \frac{-6\sqrt{2} - 8}{4} \]

### Step 5: Simplify the roots
Now we simplify both roots:
1. For \(x_1\):
\[ x_1 = \frac{-6\sqrt{2} + 8}{4} = \frac{8 - 6\sqrt{2}}{4} = 2 - \frac{3\sqrt{2}}{2} \]
2. For \(x_2\):
\[ x_2 = \frac{-6\sqrt{2} - 8}{4} = \frac{-6\sqrt{2} - 8}{4} = -2 - \frac{3\sqrt{2}}{2} \]

### Conclusion
The roots of the equation \(2x^2 + 6\sqrt{2}x + 1 = 0\) are:
\[ x_1 = 2 - \frac{3\sqrt{2}}{2}, \quad x_2 = -2 - \frac{3\sqrt{2}}{2} \]