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

Options

A. p=1
B. `p=(3)/(2)`
C. `p=2`
D. `p=-1`

Answer

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

### Step 1: Identify the coefficients
The given quadratic equation is in the form \( ax^2 + bx + c = 0 \), where:
- \( a = 2 \)
- \( b = -2(p-2) \)
- \( c = -(p+1) \)

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

### 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 we found:
\[ \alpha^2 + \beta^2 = (p - 2)^2 - 2\left(-\frac{(p+1)}{2}\right) \]
This simplifies to:
\[ \alpha^2 + \beta^2 = (p - 2)^2 + (p + 1) \]

### Step 4: Expand and simplify the expression
Expanding \( (p - 2)^2 \):
\[ (p - 2)^2 = p^2 - 4p + 4 \]
Thus,
\[ \alpha^2 + \beta^2 = p^2 - 4p + 4 + p + 1 = p^2 - 3p + 5 \]

### Step 5: Find the minimum value of the quadratic function
To find the minimum value of the quadratic function \( f(p) = p^2 - 3p + 5 \), we can use the vertex formula:
The vertex of a parabola given by \( ax^2 + bx + c \) occurs at \( p = -\frac{b}{2a} \).
Here, \( a = 1 \) and \( b = -3 \):
\[ p = -\frac{-3}{2 \cdot 1} = \frac{3}{2} \]

### Step 6: Confirm it is a minimum
To confirm that this point is indeed a minimum, we can check the second derivative:
\[ f''(p) = 2 \]
Since \( f''(p) > 0 \), the function has a minimum at \( p = \frac{3}{2} \).

### Conclusion
The value of \( p \) for which the sum of the squares of the roots is least is:
\[ \boxed{\frac{3}{2}} \]