Question

If the sum of the squares of the roots of the equation \[ x^2 - (a-2)x - (a+1) = 0 \] is least for an appropriate real parameter \( a \), then the value of \( a \) will be:

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

Hint:

To minimize root expressions: \begin{itemize} \item Convert into quadratic in parameter. \item Complete the square. \end{itemize}

Answer 1

The Correct Option is C

Concept: Let roots be \( \alpha, \beta \). 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.} \[ S = (a-2)^2 - 2[-(a+1)] \] \[ = a^2 - 4a + 4 + 2a + 2 \] \[ = a^2 - 2a + 6 \] Step 2: {\color{red}Minimize expression.} \[ S(a) = a^2 - 2a + 6 \] Complete square: \[ = (a-1)^2 + 5 \] Minimum occurs at: \[ a = 1 \]