Question

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

• A. \( \frac{(\alpha - \beta)^2}{4} \)
• B. \( (\alpha + \beta)^2 - \alpha\beta \)
• C. \( (\alpha + \beta)^2 + \alpha\beta \)
• D. \( \frac{(\alpha - \beta)^2}{2} + \alpha\beta \)
• E. \( \frac{(\alpha + \beta)^2}{2} + \alpha\beta \)

Hint:

For any quadratic $ax^2 + bx + c = 0$, the discriminant $\mathcal{D} = b^2 - 4ac$ is always equal to $a^2(\alpha - \beta)^2$.

Answer 1

The Correct Option is A

Concept: We use Vieta's formulas for the sum and product of roots:
• \( \alpha + \beta = -2b \)
• \( \alpha\beta = c \) We then express the required quantity \( b^2 - c \) in terms of \( \alpha \) and \( \beta \).

Step 1: Express \( b \) and \( c \) in terms of roots.
From the sum of roots: \( b = -\frac{\alpha + \beta}{2} \). From the product of roots: \( c = \alpha\beta \).

Step 2: Calculate \( b^2 - c \).
\[ b^2 - c = \left( -\frac{\alpha + \beta}{2} \right)^2 - \alpha\beta \] \[ b^2 - c = \frac{(\alpha + \beta)^2}{4} - \alpha\beta \] \[ b^2 - c = \frac{(\alpha + \beta)^2 - 4\alpha\beta}{4} \]

Step 3: Simplify using algebraic identities.
Recall that \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \). \[ b^2 - c = \frac{(\alpha - \beta)^2}{4} \]