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 December 13th, 2015, 01:52 AM #1 Newbie   Joined: Dec 2015 From: canada Posts: 1 Thanks: 0 Quadratic with odd coefficients Have no clue where to start a thread - but I ended up here where I can actually write something so i'm gonna just post my question I need help with! Thanks and please respond. Suppose that a quadratic P(x) has all odd coefficients. Prove that P(x) = 0 has no rational roots. Polynomial: is not given. Above is the exact given question. December 13th, 2015, 02:41 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 The quadratic can be written as (ax + p)(bx + q) = abx² + (aq + bp)x + pq, where a, b, p and q are integers, if there are rational roots. If the coefficients ab and pq are odd, a, b, p and q must all be odd. It follows that aq and bp are odd, and so the coefficient aq + bp is even. Hence there cannot be rational roots if all the coefficients are odd. Thanks from Country Boy Tags algebra, coefficients, finder, odd, quadratic, rule ### quadratic equation with odd coefficients

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