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

Options

A. 1
B. `(3)/(2)`
C. 2
D. `(5)/(2)`

Answer

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

### Step 1: Identify the coefficients
The given quadratic equation is:
\[ 2x^2 - 2(k-2)x - (k+1) = 0 \]
Here, \( a = 2 \), \( b = -2(k-2) \), and \( c = -(k+1) \).

### Step 2: Use the formulas for the sum and product of the roots
The sum of the roots \( \alpha + \beta \) is given by:
\[ \alpha + \beta = -\frac{b}{a} = -\frac{-2(k-2)}{2} = k - 2 \]
The product of the roots \( \alpha \beta \) is given by:
\[ \alpha \beta = \frac{c}{a} = \frac{-(k+1)}{2} \]

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

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

### Step 5: Minimize the quadratic expression
To minimize \( k^2 - 3k + 5 \), we can use calculus or complete the square. We will differentiate this expression:
\[ \frac{d}{dk}(k^2 - 3k + 5) = 2k - 3 \]
Setting the derivative to zero to find critical points:
\[ 2k - 3 = 0 \implies k = \frac{3}{2} \]

### Step 6: Verify that this is a minimum
To confirm that this critical point is indeed a minimum, we can check the second derivative:
\[ \frac{d^2}{dk^2}(k^2 - 3k + 5) = 2 \]
Since the second derivative is positive, the function has a minimum at \( k = \frac{3}{2} \).

### Conclusion
Thus, the value of \( k \) for which the sum of the squares of the roots is minimized is:
\[ \boxed{\frac{3}{2}} \]