The value of p for which the sum of the squares of the roots of the equation `x^(2)-(p-2)x-p-1=0` assume the least value is :

Options

A. `-1`
B. 1
C. 0
D. 2

Answer

To find the value of \( p \) for which the sum of the squares of the roots of the equation \( x^2 - (p-2)x - (p+1) = 0 \) assumes the least value, we can follow these steps:

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

### Step 2: Calculate the sum and product of the roots
Using Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = p - 2 \)
- The product of the roots \( \alpha \beta = \frac{c}{a} = -(p + 1) \)

### Step 3: Express the sum of the squares of the roots
The sum of the squares of the roots can be expressed as:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values from Step 2:
\[ \alpha^2 + \beta^2 = (p - 2)^2 - 2(-p - 1) \]
This simplifies to:
\[ \alpha^2 + \beta^2 = (p - 2)^2 + 2p + 2 \]

### Step 4: Expand and simplify
Now, we expand \( (p - 2)^2 \):
\[ (p - 2)^2 = p^2 - 4p + 4 \]
Thus, we have:
\[ \alpha^2 + \beta^2 = p^2 - 4p + 4 + 2p + 2 \]
Combining like terms gives:
\[ \alpha^2 + \beta^2 = p^2 - 2p + 6 \]

### Step 5: Find the minimum value
To find the minimum value of \( p^2 - 2p + 6 \), we can complete the square:
\[ p^2 - 2p + 6 = (p - 1)^2 + 5 \]
The expression \( (p - 1)^2 \) is always non-negative and achieves its minimum value of 0 when \( p = 1 \).

### Conclusion
Thus, the value of \( p \) for which the sum of the squares of the roots assumes the least value is:
\[ \boxed{1} \]