Find the number of integral values of p if the roots of the equation `2x^2+p^2-(5x+6p)+8=0` are of opposite sign.

Options

A. 1
B. 2
C. 3
D. 6

Answer

To find the number of integral values of \( p \) for which the roots of the equation

\[ 2x^2 + p^2 - 5x + 6p + 8 = 0 \]

are of opposite signs, we can follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation can be rewritten in the standard form \( ax^2 + bx + c = 0 \). Here:
- \( a = 2 \)
- \( b = -5 \)
- \( c = p^2 + 6p + 8 \)

### Step 2: Condition for opposite signs
For the roots \( \alpha \) and \( \beta \) to be of opposite signs, the product of the roots must be negative, i.e.,

\[ \alpha \beta < 0 \]

From Vieta's formulas, we know that

\[ \alpha \beta = \frac{c}{a} = \frac{p^2 + 6p + 8}{2} \]

Thus, the condition becomes:

\[ \frac{p^2 + 6p + 8}{2} < 0 \]

### Step 3: Simplify the inequality
Multiplying both sides by 2 (since 2 is positive, the inequality remains unchanged):

\[ p^2 + 6p + 8 < 0 \]

### Step 4: Factor the quadratic expression
Next, we need to factor the quadratic expression \( p^2 + 6p + 8 \). We can find the roots using the quadratic formula:

\[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} = \frac{-6 \pm \sqrt{36 - 32}}{2} = \frac{-6 \pm 2}{2} \]

Calculating the roots:

\[ p = \frac{-4}{2} = -2 \quad \text{and} \quad p = \frac{-8}{2} = -4 \]

### Step 5: Analyze the intervals
The roots of the quadratic \( p^2 + 6p + 8 \) are \( p = -2 \) and \( p = -4 \). The quadratic opens upwards (since the coefficient of \( p^2 \) is positive). Thus, the expression \( p^2 + 6p + 8 < 0 \) is satisfied between the roots:

\[ -4 < p < -2 \]

### Step 6: Identify integral values
Now we need to find the integral values of \( p \) in the interval \( (-4, -2) \). The only integer in this interval is:

\[ p = -3 \]

### Conclusion
Thus, the number of integral values of \( p \) such that the roots of the equation are of opposite signs is:

\[ \boxed{1} \]