Question

If \( \alpha \) and \( \beta \) are the roots of \( x^2 - ax - b = 0 \), and \( \alpha^2 + \beta^2 \) and \( \alpha^3 + \beta^3 \) are the roots of \( Ax^2 + Bx + C = 0 \), then \( C = \ldots \)

• A. \( a^5 - 5a^1b + 6ab^2 \)
• B. \( a^5 + 5a^1b - 6ab^2 \)
• C. \( a^5 - 5a^1b - 6ab^2 \)
• D. \( a^5 + 5a^1b + 6ab^2 \)

Hint:

Use identities for sums and cubes of roots to simplify the expressions. Always relate them back to the coefficients using Vieta's formulas.

Answer 1

The Correct Option is A

From Vieta’s formulas, for the equation \( x^2 - ax - b = 0 \), we have: \[ \alpha + \beta = a, \quad \alpha \beta = -b \] Now, using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \), we get: \[ \alpha^2 + \beta^2 = a^2 + 2b \] Similarly, for \( \alpha^3 + \beta^3 \), we use the identity: \[ \alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha \beta) \] This simplifies to: \[ \alpha^3 + \beta^3 = a(a^2 + 3b) \] Step 1: The sum of the roots for \( Ax^2 + Bx + C = 0 \) is: \[ \alpha^2 + \beta^2 + \alpha^3 + \beta^3 = -\frac{B}{A} \] Step 2: The product of the roots is: \[ (\alpha^2 + \beta^2)(\alpha^3 + \beta^3) = \frac{C}{A} \] After simplifying, we find that: \[ C = a^5 - 5a^1b + 6ab^2 \] % Final Answer \[ \boxed{a^5 - 5a^1b + 6ab^2} \]