Question

If the roots of equation x³ + ax² + b = 0 are α₁, α₂ and α₃;

(a, b ≠ 0) then the equation whose roots are $\frac{\alpha_{1}\alpha_{2} + \alpha_{2}\alpha_{3}}{\alpha_{1}\alpha_{2}\alpha_{3}}$,$\frac{\alpha_{2}\alpha_{3} + \alpha_{3}\alpha_{1}}{\alpha_{1}\alpha_{2}\alpha_{3}}$,$\frac{\alpha_{1}\alpha_{3} + \alpha_{1}\alpha_{2}}{\alpha_{1}\alpha_{2}\alpha_{3}}$ are

• A. ax³ + bx – 1 = 0
• B. bx³ + ax – 1 = 0
• C. ax³ – bx – 1 = 0
• D. bx³ + ax + 1 = 0

Answer 1

Answer: (B)

bx³ + ax – 1 = 0

Answer 2

$\frac{\alpha_{1}\alpha_{2} + \alpha_{2}\alpha_{3}}{\alpha_{1}\alpha_{2}\alpha_{3}}$= $–\frac{\alpha_{1}\alpha_{3}}{\alpha_{1}\alpha_{2}\alpha_{3}}$ = $–\frac{1}{\alpha_{2}}$.

**∴ **required equation has root $–\frac{1}{\alpha_{1}},–\frac{1}{\alpha_{2}},–\frac{1}{\alpha_{3}}$

⇒ y = $–\frac{1}{x}$ or x = $–\frac{1}{y}$.

∴ required equation$\left( –\frac{1}{y} \right)^{3}$+ a$\left( –\frac{1}{y} \right)^{2}$+ b = 0

Or by³ + ay – 1 = 0

⇒ bx³ + ax – 1 = 0