If the roots of the equation `x^(2)+5x-p=0` differ by unity, then the value of p is

Options

A. `-6`
B. `-5`
C. 6
D. 12

Answer

To solve the problem, we need to find the value of \( p \) such that the roots of the quadratic equation \( x^2 + 5x - p = 0 \) differ by 1.

### Step 1: Identify the coefficients
The given quadratic equation is:
\[ x^2 + 5x - p = 0 \]
Here, we can identify the coefficients:
- \( a = 1 \)
- \( b = 5 \)
- \( c = -p \)

### Step 2: Use the relationship between the roots
Let the roots of the equation be \( \alpha \) and \( \beta \). According to Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -\frac{5}{1} = -5 \)
- The product of the roots \( \alpha \beta = \frac{c}{a} = \frac{-p}{1} = -p \)

### Step 3: Set up the equation based on the difference of roots
We are given that the roots differ by 1, which means:
\[ |\alpha - \beta| = 1 \]
This can be expressed as:
\[ \alpha - \beta = 1 \quad \text{or} \quad \beta - \alpha = 1 \]
Without loss of generality, we can take:
\[ \alpha - \beta = 1 \]

### Step 4: Express \( \alpha \) in terms of \( \beta \)
From the equation \( \alpha - \beta = 1 \), we can express \( \alpha \) as:
\[ \alpha = \beta + 1 \]

### Step 5: Substitute \( \alpha \) in the sum of roots
Now, substituting \( \alpha \) in the sum of roots:
\[ (\beta + 1) + \beta = -5 \]
This simplifies to:
\[ 2\beta + 1 = -5 \]
Subtracting 1 from both sides gives:
\[ 2\beta = -6 \]
Dividing by 2 results in:
\[ \beta = -3 \]

### Step 6: Find \( \alpha \)
Now substituting back to find \( \alpha \):
\[ \alpha = \beta + 1 = -3 + 1 = -2 \]

### Step 7: Calculate the product of the roots
Now we can find the product of the roots:
\[ \alpha \beta = (-2)(-3) = 6 \]
From Vieta's formulas, we know:
\[ \alpha \beta = -p \]
Thus:
\[ 6 = -p \implies p = -6 \]

### Final Answer
The value of \( p \) is:
\[ \boxed{-6} \]