The value of p for which the difference between the roots of the equation `x^(2) + px + 8 = 0` is 2, are

Options

A. `+-2`
B. `+-4`
C. `+-6`
D. `+-8`

Answer

To find the value of \( p \) for which the difference between the roots of the equation \( x^2 + px + 8 = 0 \) is 2, we can follow these steps:

### Step 1: Define the roots
Let the roots of the quadratic equation be \( \alpha \) and \( \beta \). According to the problem, the difference between the roots is given by:
\[ \alpha - \beta = 2 \]

### Step 2: Use Vieta's Formulas
From Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -p \)
- The product of the roots \( \alpha \beta = \frac{c}{a} = 8 \)

### Step 3: Set up equations
We have the following equations:
1. \( \alpha + \beta = -p \) (Equation 1)
2. \( \alpha - \beta = 2 \) (Equation 2)
3. \( \alpha \beta = 8 \) (Equation 3)

### Step 4: Solve for \( \alpha \) and \( \beta \)
From Equation 2, we can express \( \alpha \) in terms of \( \beta \):
\[ \alpha = \beta + 2 \]

Now, substitute \( \alpha \) in Equation 1:
\[ (\beta + 2) + \beta = -p \]
This simplifies to:
\[ 2\beta + 2 = -p \]
So,
\[ 2\beta = -p - 2 \quad \Rightarrow \quad \beta = \frac{-p - 2}{2} \]

Now substitute \( \beta \) back into the expression for \( \alpha \):
\[ \alpha = \frac{-p - 2}{2} + 2 = \frac{-p - 2 + 4}{2} = \frac{-p + 2}{2} \]

### Step 5: Substitute into the product equation
Now, substitute \( \alpha \) and \( \beta \) into Equation 3:
\[ \left(\frac{-p + 2}{2}\right) \left(\frac{-p - 2}{2}\right) = 8 \]
This simplifies to:
\[ \frac{(-p + 2)(-p - 2)}{4} = 8 \]
Multiplying both sides by 4 gives:
\[ (-p + 2)(-p - 2) = 32 \]

### Step 6: Expand and rearrange
Expanding the left side:
\[ (p^2 - 4) = 32 \]
Rearranging gives:
\[ p^2 - 36 = 0 \]

### Step 7: Solve for \( p \)
Factoring the quadratic:
\[ (p - 6)(p + 6) = 0 \]
Thus, we find:
\[ p = 6 \quad \text{or} \quad p = -6 \]

### Final Answer
The values of \( p \) for which the difference between the roots of the equation is 2 are:
\[ p = 6 \quad \text{and} \quad p = -6 \]

---