# algebra

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If p≠q and p² = 5p-3 and q² = 5q-3 the equation having roots as p/q and q/p is

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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) = (19-5√13)/6

So, the equation is

(x-((19+5√13)/6))(x-((19-5√13)/6)) = 0
(6x-19-5√13)(6x-19+5√13) = 0
(6x-19)^2 - 325 = 0
36x^2 - 228x + 36 = 0
3x^2 - 19x + 3 = 0

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