Let `p, q in R`. If `2- sqrt3` is a root of the quadratic equation, `x^(2)+px+q=0,` then

Options

A. `q^(2)-4p-16=0`
B. `p^(2)-4q+12=0`
C. `p^(2)-4q-12=0`
D. `q^(2)+4p+14=0`

Explanation

To solve the problem, we need to determine the values of \( p \) and \( q \) given that \( 2 - \sqrt{3} \) is a root of the quadratic equation \( x^2 + px + q = 0 \).

### Step-by-Step Solution:

1. **Identify the Roots**:
Since \( 2 - \sqrt{3} \) is a root of the quadratic equation, we can use the property of quadratic equations that states if \( \alpha + \sqrt{\beta} \) is one root, then the other root is \( \alpha - \sqrt{\beta} \).
Here, \( \alpha = 2 \) and \( \beta = 3 \).
Therefore, the other root is:
\[ 2 + \sqrt{3} \]

2. **Calculate the Sum of Roots**:
The sum of the roots of a quadratic equation \( x^2 + px + q = 0 \) is given by:
\[ \text{Sum of roots} = -\frac{p}{1} = -p \]
The sum of our roots \( (2 - \sqrt{3}) + (2 + \sqrt{3}) \) simplifies to:
\[ 4 \]
Thus, we have:
\[ -p = 4 \implies p = -4 \]

3. **Calculate the Product of Roots**:
The product of the roots is given by:
\[ \text{Product of roots} = \frac{q}{1} = q \]
The product of our roots \( (2 - \sqrt{3})(2 + \sqrt{3}) \) can be calculated using the difference of squares:
\[ (2)^2 - (\sqrt{3})^2 = 4 - 3 = 1 \]
Therefore, we have:
\[ q = 1 \]

4. **Final Values**:
We have determined:
\[ p = -4 \quad \text{and} \quad q = 1 \]

### Conclusion:
The values of \( p \) and \( q \) are:
\[ p = -4, \quad q = 1 \]