algebra
posted by disha .
If p≠q and p² = 5p3 and q² = 5q3 the equation having roots as p/q and q/p is

well
p = (5±√13)/2
q = (5±√13)/2
If p≠q then one is (5+√13)/2 and the other is (5√13)/2
(5+√13)/(5√13) = (19+5√13)/6
(5√13)/(5+√13) = (195√13)/6
So, the equation is
(x((19+5√13)/6))(x((195√13)/6)) = 0
(6x195√13)(6x19+5√13) = 0
(6x19)^2  325 = 0
36x^2  228x + 36 = 0
3x^2  19x + 3 = 0
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