If roots of the equation `a x^2+b x+c=0` are `alpha and beta` , find the equation whose roots are `1/alpha,1/beta` (ii) `-alpha,-beta` (iii) `(1-alpha)/(1+alpha),(1-beta)/(1+beta)`

Explanation

Here in all cases, functions of `alpha and beta` are symmetric.
(i) Let `(1)/(alpha)=y or alpha=(1)/(y)`
Now `alpha ` is a root of the equation `ax^(2) + bx + c = 0` . Thus,
`aalpha^(2) + balpha + c = 0`
or `(a)/(y^(2))+ (b)/(y) + c = 0`
or ` cy^(2) + by + a = 0`
Hence, the requried equation is `cx^(2) + bx + a = 0`
We get same equation if we start with `(1)/(beta) = y`
(ii) Let `-alpha = y or alpha = -y`
Now `alpha ` is root of the equation `ax^(2) + bx + c = 0`. Thus ,
`aalpha^(2)+ balpha + c = 0`
Hence, the required is `ax^(2) - bx + c = 0`
(iii) Let `(1-alpha)/(1+alpha) = y or alpha = (1-y)/(1+y)`
Now `alpha` is root of the equation `ax^(2) + bx + c = 0.`Thus ,
`aalpha^(2) + balpha + c = 0`
or `a((1-y)/(1+y))^(2) +b((1-y)/(1+y)) + c=0`
Hence, required equation is
`a(1 - x)^(2) + b (1 - x^(2)) + c (1 + x)^(2) = 0`