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 will follow these steps:

### Step 1: Identify the roots and their properties
The roots \( x_1 \) and \( x_2 \) of the quadratic equation \( x^2 + px + 12 = 0 \) have the following properties:
- The sum of the roots \( x_1 + x_2 = -\frac{b}{a} = -p \)
- The product of the roots \( x_1 x_2 = \frac{c}{a} = 12 \)

### Step 2: Set up the equations
From the properties of the roots, we can write:
1. \( x_1 + x_2 = -p \) (1)
2. \( x_1 x_2 = 12 \) (2)

We are also given that the difference between the roots is 1:
\[ |x_1 - x_2| = 1 \]

### Step 3: Express the difference of the roots
Assuming \( x_1 - x_2 = 1 \) (without loss of generality), we can express \( x_1 \) in terms of \( x_2 \):
\[ x_1 = x_2 + 1 \]

### Step 4: Substitute \( x_1 \) in the sum and product equations
Substituting \( x_1 \) in equation (1):
\[ (x_2 + 1) + x_2 = -p \]
\[ 2x_2 + 1 = -p \]
\[ 2x_2 = -p - 1 \]
\[ x_2 = -\frac{p + 1}{2} \] (3)

Now substituting \( x_1 \) and \( x_2 \) in equation (2):
\[ (x_2 + 1)x_2 = 12 \]
Substituting \( x_2 \) from equation (3):
\[ \left(-\frac{p + 1}{2} + 1\right)\left(-\frac{p + 1}{2}\right) = 12 \]
\[ \left(-\frac{p - 1}{2}\right)\left(-\frac{p + 1}{2}\right) = 12 \]
\[ \frac{(p - 1)(p + 1)}{4} = 12 \]

### Step 5: Solve for \( p \)
Multiplying both sides by 4:
\[ (p - 1)(p + 1) = 48 \]
Expanding the left side:
\[ p^2 - 1 = 48 \]
\[ p^2 = 49 \]
\[ p = 7 \quad \text{or} \quad p = -7 \]

Since we are looking for the positive value of \( p \):
\[ p = 7 \]

### Final Answer
The positive value of \( p \) is \( 7 \).

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