For t > −1, let αₜ and βₜ be the roots of the equation ((t + 2)¹/⁶ − 1)x² +((t + 2)¹/⁶ − 1)x + ((t + 2)¹/²¹ − 1) = 0. If lim t → −1⁺ αₜ = a and lim t → −1⁺ βₜ = b, then 72(a + b)² is equal to:
❓ Question: For t > − 1 t > -1 , let α t \alpha_t and β t \beta_t be the roots of the equation: ( ( t + 2 ) 1 / 6 − 1 ) x 2 + ( ( t + 2 ) 1 / 6 − 1 ) x + ( ( t + 2 ) 1 / 21 − 1 ) = 0. If lim t → − 1 + α t = a and lim t → − 1 + β t = b , then find 72 ( a + b ) 2 . 🖼️ Question Image ✍️ Short Solution Write the quadratic in standard form: A ( t ) x 2 + B ( t ) x + C ( t ) = 0 where A ( t ) = ( t + 2 ) 1 / 6 − 1 , B ( t ) = ( t + 2 ) 1 / 6 − 1 , C ( t ) = ( t + 2 ) 1 / 21 − 1. Observe the limits as t → − 1 + t \to -1^+ : A ( t ) → ( 1 ) − 1 = 0 , B ( t ) → 0 , C ( t ) → 2 1 / 21 − 1 but we must use L’Hospital-type analysis since coefficients vanish. Step 1: Factor out A ( t ) A(t) from the quadratic x 2 + x + C ( t ) A ( t ) = 0 ⇒ x 2 + x + ( t + 2 ) 1 / 21 − 1 ( t + 2 ) 1 / 6 − 1 = 0. Step 2: Evaluate the limit Set h = t + 2 h = t+2 , then as t → − 1 + t\to -1^+ , h → 1 + h\to 1^+ . Then: ( t + 2 ) 1 / 21 − 1 ( t + 2 ) 1 / 6 − 1 = h ...