Question

If the expression \( 7 + 6x - 3x^2 \) attains its extreme value \( \beta \) at \( x = \alpha \), then the sum of the squares of the roots of the equation \( x^2 + ax - \beta = 0 \) is:

• A. \( 21 \)
• B. \( -19 \)
• C. \( 19 \)
• D. -21

Hint:

To find the sum of the squares of the roots, use the identity \( \alpha^2 + \beta^2 = ( \alpha + \beta)^2 - 2\alpha \beta \).

Answer 1

The Correct Option is A

Step 1: The given expression is \( f(x) = 7 + 6x - 3x^2 \). To find the extreme value, we begin by differentiating: \[ f'(x) = 6 - 6x. \] Setting \( f'(x) = 0 \), we solve for \( x \): \[ 6 - 6x = 0 \quad \Rightarrow \quad x = 1. \] Therefore, \( \alpha = 1 \). 

Step 2: Determine the extreme value \( \beta \). Substitute \( x = 1 \) into the expression for \( f(x) \): \[ f(1) = 7 + 6(1) - 3(1)^2 = 7 + 6 - 3 = 10. \] Thus, \( \beta = 10 \). 

Step 3: The equation is \( x^2 + ax - \beta = 0 \), where \( \beta = 10 \). Using Vieta's formulas, the sum of the squares of the roots is given by: \[ \text{Sum of squares of roots} = (\text{Sum of roots})^2 - 2 \times \text{Product of roots}. \] For this quadratic, the sum of the roots is \( -a \) and the product is \( -\beta = -10 \). Therefore, the sum of the squares of the roots is \( 21 \).