Question

solve ques no. 9 .. fast .. plz

Answer 1

Hi friend!!

α and ß are the zeroes of the given polynomial.
Given polynomial, 6x²-x+1

Sum of zeroes = 1/6 = -x coefficient/x² coefficient
α and ß = 1/6

Product of zeroes = 1/6 = constant/x² coefficient
αß = 1/6


α/ß + ß/α + 2{1/α + 1/ß} + 3αß

→ (α²+ß²)/αß + 2(ß+α)/αß + 3αß

→ {α²+ß²+2αß - 2αß}/αß + 2{1/6÷1/6} + 3(1/6)

→ {(α+ß)² - 2(1/6)}/1/6 + 2 + ½

→ {(1/6)² - ⅓}/1/6 + 2 + ½

→ (1/36 - ⅓) × 6/1 + (4+1)/2

→ (1-12)/36 × 6 + 5/2

→ -11/6 + 5/2

→ (-11+15)/6

→ 4/6

→⅔

→α/ß + ß/α + 2{1/α + 1/ß} + 3αß = ⅔

Hope it helps....