Question

if α and βare the zeros of QP x^2+6x+9 then find α/β+β/α

Answer 1

P ( X ) = X² + 6X + 9


Here,


A = 1 , B = 6 and C = 9



Sum of zeroes = -B/A


Alpha + Beta = -6



And,

Product of zeroes = C/A


Alpha × Beta = 9



Therefore,



Alpha / Beta + Beta / Alpha = ( Alpha² + Beta² ) / Alpha × Beta



=> ( Alpha + Beta )² - 2Alphabeta / 9




=> ( -6)² - 2 × 9 / 9



=> 36 - 18 / 9



=> 18/9


=> 2

Answer 2

X² + 6X + 9


a = 1


b = 6 and C = 9


sum of zeroes = -b/a

@ + ß = -6

and,

Product of zeroes = c/a

@ × ß = 9


Therefore,

@ / ß + ß /a

= @² + ß² /@b

= 36 - 18/9

= 2