If the zero of polynomial ax square+bc+c are in ratio 4:5 prove that 2

1.	If the zero of polynomial ax square+bc+c are in ratio 4:5 prove that 20 b square=81ac
Answer» Let roots 4k and 5k4k+5k =-b/a9k =-b/aK=-b/9a..... (1)4k*5k=c/a20k2=c/aSub ... (1)20 (-b/9a)2=c/a20 (b2/81a2)=c/a20 b2\xa0= 81 ac Let the zeroes be 4p, 5p.We know that the sum of the zeroes=-b/a. Then, 4p+5p=9p=-b/a.p=-b/9a, then, p^2=b^2/81a^2...(1)And the product of zeroes is c/a.Then 4p×5p=c/a.20p^2=c/a then, p^2=c/20a...(2)Using (1) and (2).b^2/81a^2=c/20a.Then, 20b^2=81ac.

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