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 coefficients of the quadratic equation.
The given quadratic equation is \( x^2 - (a - 2)x - a + 1 = 0 \).
Comparing this with the standard form \( Ax^2 + Bx + C = 0 \), we have:
\( A = 1 \)
\( B = -(a - 2) = 2 - a \)
\( C = -(a - 1) = 1 - a \)

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

Step 3: Express the sum of the squares of the roots in terms of \( 'a' \).
We want to find the minimum value of \( \alpha^2 + \beta^2 \). We can express this in terms of the sum and product of the roots: \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \] Substitute the expressions for \( \alpha + \beta \) and \( \alpha\beta \) in terms of \( 'a' \): \[ \alpha^2 + \beta^2 = (a - 2)^2 - 2(1 - a) \] \[ \alpha^2 + \beta^2 = (a^2 - 4a + 4) - (2 - 2a) \] \[ \alpha^2 + \beta^2 = a^2 - 4a + 4 - 2 + 2a \] \[ \alpha^2 + \beta^2 = a^2 - 2a + 2 \]
Step 4: Find the value of \( 'a' \) that minimizes the sum of the squares of the roots.
We have the expression for the sum of the squares of the roots as a quadratic function of \( 'a' \): \( S(a) = a^2 - 2a + 2 \). To find the minimum value of this quadratic function, we can complete the square or use calculus. Completing the square: \[ S(a) = (a^2 - 2a + 1) + 1 \] \[ S(a) = (a - 1)^2 + 1 \] Since \( (a - 1)^2 \ge 0 \) for all real values of \( 'a' \), the minimum value of \( S(a) \) occurs when \( (a - 1)^2 = 0 \), which means \( a - 1 = 0 \), so \( a = 1 \). The minimum value of the sum of the squares of the roots is \( 0 + 1 = 1 \). Alternatively, using calculus, we can find the critical points by taking the derivative of \( S(a) \) with respect to \( 'a' \) and setting it to zero: \[ S'(a) = \frac{d}{da}(a^2 - 2a + 2) = 2a - 2 \] Set \( S'(a) = 0 \): \[ 2a - 2 = 0 \implies 2a = 2 \implies a = 1 \] To confirm that this is a minimum, we can check the second derivative: \[ S''(a) = \frac{d}{da}(2a - 2) = 2 \] Since \( S''(a) = 2>0 \), the function \( S(a) \) has a minimum at \( a = 1 \). Therefore, the sum of the squares of the roots will have the least value when \( a = 1 \).