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:

Remember Vieta's formulas for relating the coefficients of a polynomial to sums and products of its roots. For a quadratic equation \( Ax^2 + Bx + C = 0 \), the sum of the roots is \( -\frac{B}{A} \) and the product of the roots is \( \frac{C}{A} \).

Answer 1

The Correct Option is D

Step 1: Identify the equation's coefficients.
\nThe given equation is \( x^2 - (a - 2)x - a + 1 = 0 \).
\nComparing with \( Ax^2 + Bx + C = 0 \), we get:
\n\( A = 1 \)
\n\( B = -(a - 2) = 2 - a \)
\n\( C = -(a - 1) = 1 - a \)
\n\n
Step 2: Apply Vieta's formulas for sum and product of roots.
\nLet the roots be \( \alpha \) and \( \beta \). Vieta's formulas give us:
\nSum: \( \alpha + \beta = -\frac{B}{A} = - \frac{(2 - a)}{1} = a - 2 \)
\nProduct: \( \alpha \beta = \frac{C}{A} = \frac{(1 - a)}{1} = 1 - a \)
\n\n
Step 3: Express sum of squares in terms of \( 'a' \).
\nWe want to minimize \( \alpha^2 + \beta^2 \). Express this using sum and product:\n\[\n\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\n\]\nSubstitute the expressions in terms of \( 'a' \):\n\[\n\alpha^2 + \beta^2 = (a - 2)^2 - 2(1 - a)\n\]\n\[\n\alpha^2 + \beta^2 = (a^2 - 4a + 4) - (2 - 2a)\n\]\n\[\n\alpha^2 + \beta^2 = a^2 - 4a + 4 - 2 + 2a\n\]\n\[\n\alpha^2 + \beta^2 = a^2 - 2a + 2\n\]\n\n
Step 4: Find the value of \( 'a' \) that minimizes the sum of the squares.
\nWe have \( S(a) = a^2 - 2a + 2 \). To minimize, complete the square or use calculus.\n\nCompleting the square:\n\[\nS(a) = (a^2 - 2a + 1) + 1\n\]\n\[\nS(a) = (a - 1)^2 + 1\n\]\nSince \( (a - 1)^2 \ge 0 \), the minimum occurs when \( (a - 1)^2 = 0 \), so \( a = 1 \).\nThe minimum sum of squares is \( 0 + 1 = 1 \).\n\nAlternatively, using calculus, find critical points by setting the derivative to zero:\n\[\nS'(a) = \frac{d}{da}(a^2 - 2a + 2) = 2a - 2\n\]\nSet \( S'(a) = 0 \):\n\[\n2a - 2 = 0 \implies 2a = 2 \implies a = 1\n\]\nCheck the second derivative:\n\[\nS''(a) = \frac{d}{da}(2a - 2) = 2\n\]\nSince \( S''(a) = 2>0 \), \( S(a) \) has a minimum at \( a = 1 \).\n\nTherefore, the sum of squares is minimized when \( a = 1 \).