-2 and 3 are the roots of ax^2 + bx + c = 0. What are the roots of cx^2 + bx + a?
The sum of the roots = -b/a
So 3 - 2 = -b/a = 1 which implies that -a = b
The product of the roots = c/a
So (-2)(3) = c/a
-6 = c/a
-6a = c
In the second polynomial we have that
-6ax^2 - ax + a = 0 divide through by a
-6x^2 - x + 1 = 0
6x^2 + x - 1 = 0
(3x -1) (2x + 1) = 0
Set both factors to 0 and solve for x and we have that
x = 1/3 and x = -1/2
Just the reciprocals of the original roots....that's a neat result !!!!!