Question

If α and β are roots of the equation x² – 11x + 28 = 0, then the equation whose roots are α2/β and β2/α is:
1. 28x² – 407x + 784 = 0
2. 28x² – 407x + 284 = 0
3. 28x² – 347x + 784 = 0
4. 40x² – 407x + 784 = 0

Answer 1

Correct Answer - Option 1 : 28x² – 407x + 784 = 0

Given:

x² – 11x + 28 = 0

Concept Used:

Equation whose roots are α²/β and β²/α:

x² - ( sum of the roots) x + product of the roots = 0

⇒ x² – (α²/β + β²/α)x + (α²/β × β²/α) = 0

⇒ x² – (α²/β + β²/α)x + (αβ) = 0

Calculation:

x² – 11x + 28 = 0

⇒ x² – 7x – 4x + 28 = 0

⇒ (x – 7)(x - 4) = 0

⇒ x = 4, 7

∴ α = 4 and β = 7

(α + β) = 11

⇒ α³ + β³ + 3αβ(α + β) = 1331

⇒ α³+ β³ + 84(11) = 1331

⇒ α³+ β³ + 924 = 1331

⇒ α³+ β³= 1331 – 924

⇒ α³+ β³ = 407

&, αβ = 28

Sum of roots = α²/β + β²/α = (α³ +β³)/αβ

Equation whose roots are α²/β and β²/α:

x² – (α²/β + β²/α)x + (αβ) = 0

x2 – (α2/β + β2/α)x + (αβ) = 0

⇒ x² – [(α³ +β³)/αβ]x + αβ = 0

⇒ x² – (407/28)x + 28 = 0

⇒ 28x² – 407x + 784 = 0