Question

If \( \alpha \) and \( \beta \) are the two roots of the equation \( x^2 + ax + b = 0 \), \( ab \neq 0 \), then the quadratic equation with roots \( \frac{1}{\alpha^3 + \alpha} \) and \( \frac{1}{\beta^3 + \beta} \) is

• A. \( b(b^2 + 1 + a^2 - 2b)x^2 + (a^3 + a - 3ab)x + 1 = 0 \)
• B. \( b(b^2 + 1 + a^2 + 2b)x^2 + (a^3 - a - 3ab)x + 1 = 0 \)
• C. \( b(b^2 + 1 + a^2 + 2b)x^2 + (a^3 + a - 3ab)x + 1 = 0 \)
• D. \( b(b^2 + 1 + a^2 - 2b)x^2 + (a^3 + a - 3ab)x + 1 = 0 \)

Hint:

To find a quadratic equation with roots in the form \( \frac{1}{\alpha^3 + \alpha} \) and \( \frac{1}{\beta^3 + \beta} \), use the relationships between the roots and coefficients of the original equation and apply Vieta's formulas.

Answer 1

The Correct Option is A

Step 1: Given the quadratic equation \( x^2 + ax + b = 0 \) with roots \( \alpha \) and \( \beta \), by Vieta's formulas, we know: \[ \alpha + \beta = -a \text{and} \alpha \beta = b \] 

Step 2: The new quadratic equation will have roots \( \frac{1}{\alpha^3 + \alpha} \) and \( \frac{1}{\beta^3 + \beta} \). Using the fact that the sum and product of the roots of a quadratic equation \( x^2 + px + q = 0 \) are given by \( -p \) and \( q \) respectively, we can derive the quadratic equation. 

Step 3: After deriving the equation using the relationships between the roots and coefficients, we find that the correct quadratic equation is: \[ b(b^2 + 1 + a^2 - 2b)x^2 + (a^3 + a - 3ab)x + 1 = 0 \]