The value of a so that the sum of the squares of roots of the equation `x^2-(a-2)x-a+1=0` assume the least value is

Options

A. 0
B. 1
C. 2
D. 3

Answer

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

### Step 1: Identify the coefficients
The given quadratic equation is:
\[ x^2 - (a-2)x - (a+1) = 0 \]
From this, we can identify:
- \( A = 1 \)
- \( B = -(a-2) \)
- \( C = -(a+1) \)

### Step 2: Use the relationships of roots
Let \( \alpha \) and \( \beta \) be the roots of the equation. According to Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{B}{A} = a - 2 \)
- The product of the roots \( \alpha \beta = \frac{C}{A} = -(a + 1) \)

### Step 3: Find 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 Vieta's formulas:
\[ \alpha^2 + \beta^2 = (a - 2)^2 - 2(- (a + 1)) \]

### Step 4: Simplify the expression
Now, let's simplify the expression:
\[ \alpha^2 + \beta^2 = (a - 2)^2 + 2(a + 1) \]
Expanding \( (a - 2)^2 \):
\[ = a^2 - 4a + 4 + 2a + 2 \]
Combining like terms:
\[ = a^2 - 2a + 6 \]

### Step 5: Find the minimum value
To find the value of \( a \) that minimizes \( a^2 - 2a + 6 \), we can complete the square:
\[ = (a - 1)^2 + 5 \]
The minimum value occurs when \( (a - 1)^2 = 0 \), which gives:
\[ a - 1 = 0 \Rightarrow a = 1 \]

### Conclusion
Thus, the value of \( a \) such that the sum of the squares of the roots assumes the least value is:
\[ \boxed{1} \]