Question

If \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 + \alpha x + \beta = 0 \), then

• A. \( \alpha = -1, \beta = -2 \)
• B. \( \alpha = 0, \beta = 1 \)
• C. \( \alpha = -2, \beta = 0 \)
• D. \( \alpha = -2, \beta = 1 \)
• E. \( \alpha = 1, \beta = -2 \)

Hint:

When coefficients themselves are roots, directly apply Vieta's formulas and solve the resulting system carefully.

Answer 1

The Correct Option is D

Concept: If \( \alpha \) and \( \beta \) are roots of a quadratic equation, then: \[ \text{Sum of roots} = \alpha + \beta = -(\text{coefficient of } x) \] \[ \text{Product of roots} = \alpha \beta = \text{constant term} \]

Step 1: Apply Vieta's formulas. Given equation: \[ x^2 + \alpha x + \beta = 0 \] So: \[ \alpha + \beta = -\alpha \cdots (1) \] \[ \alpha \beta = \beta \cdots (2) \]

Step 2: Solve the equations. From (2): \[ \alpha \beta = \beta \Rightarrow \beta(\alpha - 1) = 0 \] So either: \[ \beta = 0 \text{or} \alpha = 1 \] Case 1: \( \beta = 0 \) From (1): \[ \alpha + 0 = -\alpha \Rightarrow 2\alpha = 0 \Rightarrow \alpha = 0 \] This gives \( (\alpha, \beta) = (0,0) \), not in options. Case 2: \( \alpha = 1 \) From (1): \[ 1 + \beta = -1 \Rightarrow \beta = -2 \]

Step 3: Final answer. \[ \alpha = 1, \beta = -2 \]