Question

For what value of \( a \), the sum of the squares of the roots of the equation \[ x^2 - (a-2)x - a + 1 = 0 \] will have the least value?

• A. \( 2 \)
• B. \( 0 \)
• C. \( 3 \)
• D. \( 1 \)

Hint:

For quadratic root expressions: \begin{itemize} \item Use identities: \[ \alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta \] \item Convert into quadratic in parameter and minimize. \end{itemize}

Answer 1

The Correct Option is D

Concept: If roots are \( \alpha, \beta \), then: \[ \alpha + \beta = a-2, \quad \alpha\beta = -a+1 \] Sum of squares: \[ \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta \] Step 1: {\color{red}Compute sum of squares.} \[ \alpha^2 + \beta^2 = (a-2)^2 - 2(-a+1) \] \[ = a^2 - 4a + 4 + 2a - 2 \] \[ = a^2 - 2a + 2 \] Step 2: {\color{red}Minimize expression.} \[ S(a) = a^2 - 2a + 2 \] Complete square: \[ = (a-1)^2 + 1 \] Minimum occurs when: \[ a = 1 \] Step 3: {\color{red}Minimum value exists.} Thus least sum of squares occurs at \( a = 1 \).