The roots `x_(1) and x_(2)` of the equation `x^(2) +px +12=0` are such that their differences is 1. then the positive value of p is

Options

A. 1
B. 2
C. 3
D. 7

Answer

To solve the problem, we need to find the positive value of \( p \) given that the roots \( x_1 \) and \( x_2 \) of the equation \( x^2 + px + 12 = 0 \) have a difference of 1.

### Step-by-Step Solution:

1. **Understand the relationship between the roots and coefficients:**
From Vieta's formulas, we know:
- The sum of the roots \( x_1 + x_2 = -p \)
- The product of the roots \( x_1 x_2 = 12 \)

2. **Set up the equation based on the difference of the roots:**
We are given that the difference between the roots is 1:
\[ |x_1 - x_2| = 1 \]
Without loss of generality, we can assume \( x_1 - x_2 = 1 \). Therefore, we can express \( x_1 \) in terms of \( x_2 \):
\[ x_1 = x_2 + 1 \]

3. **Substitute \( x_1 \) in the product equation:**
Substitute \( x_1 \) into the product equation:
\[ (x_2 + 1)x_2 = 12 \]
Expanding this gives:
\[ x_2^2 + x_2 - 12 = 0 \]

4. **Solve the quadratic equation:**
We can solve the quadratic equation \( x_2^2 + x_2 - 12 = 0 \) using the quadratic formula:
\[ x_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 1, c = -12 \):
\[ x_2 = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} \]
\[ x_2 = \frac{-1 \pm \sqrt{1 + 48}}{2} \]
\[ x_2 = \frac{-1 \pm \sqrt{49}}{2} \]
\[ x_2 = \frac{-1 \pm 7}{2} \]
This gives us two possible values for \( x_2 \):
\[ x_2 = \frac{6}{2} = 3 \quad \text{or} \quad x_2 = \frac{-8}{2} = -4 \]

5. **Find the corresponding values of \( x_1 \):**
If \( x_2 = 3 \):
\[ x_1 = x_2 + 1 = 3 + 1 = 4 \]
If \( x_2 = -4 \):
\[ x_1 = x_2 + 1 = -4 + 1 = -3 \]

6. **Calculate \( p \) using the sum of roots:**
Now, we can find \( p \) using the sum of the roots:
- For \( x_1 = 4 \) and \( x_2 = 3 \):
\[ x_1 + x_2 = 4 + 3 = 7 \implies -p = 7 \implies p = -7 \]
- For \( x_1 = -3 \) and \( x_2 = -4 \):
\[ x_1 + x_2 = -3 + (-4) = -7 \implies -p = -7 \implies p = 7 \]

7. **Conclusion:**
The positive value of \( p \) is:
\[ \boxed{7} \]