Prove that if the co-efficients of the quadratic equation \(ax^{2}+bx+c = 0\) are odd integers,the roots of the equations cannot be rational numbers.

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TopNewestLet a = 2d + 1, b = 2e + 1, c = 2f + 1 where d,e,f are integers.

Using the quadratic formula we get x = (-b +- sqrt{b^2 - 4ac})/(2a). We will show that b^2 - 4ac cannot be a square number.

Substituting our above equations (a = 2d + 1, b = 2e + 1, c = 2f + 1) into b^2 - 4ac we get 4e^2 + 4e - 16df - 8d - 8f - 3. We can rewrite this as (8)(-2df - d - f) + (4)(e)(e + 1) - 3. Since one of (e) and (e+1) must be even, it follows that 4(e)(e+1) must be divisible by 8. Therefore b^2 - 4ac is congruent to 5 modulo 8. This means that b^2 - 4ac is not a square number, as square numbers must be congruent to 0, 1, or 4 mod 8. This shows that x is irrational. – Colin Tang · 2 years, 11 months ago

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Suppose that p/q is a root of the quadratic equation and that p,q are coprime, then c is divisible by p and a is divisible by q. Hence both p and q are odd. Substituting p/q for x, we get: \(\frac{ap^2}{q^2} + \frac{p}{q} + c = 0 \). However, this is clearly impossible as both p, q are odd. Hence there are no rational roots for the equation. – Jianzhi Wang · 2 years, 11 months ago

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– Jianzhi Wang · 2 years, 11 months ago

There are 3 odd terms in the final equation. If all of them are positive, then we are done. If one is negative, then the LHS will still be an odd number (odd + odd - odd = odd != 0).Log in to reply

– Eddie The Head · 2 years, 11 months ago

Why is it impossible?Log in to reply

Never mind, I think I've figured it out. Since a/q is odd and p is odd, (a/q)(p)(p) is odd and p is odd, which makes (a/q)(p)(p) + p even. You want (a/q)(p)(p) + p to equal -cq, which is a product of two odd integers and therefore odd. But you can't have an even number equal to an odd number. – Colin Tang · 2 years, 11 months ago

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– Bogdan Simeonov · 2 years, 11 months ago

He means that if we multiply by \(q^2\), LHS will be odd and RHS =0, clearly a contradiction.Log in to reply

integers, they can be negative, right? We could somehow choose the coefficients so that LHS is zero. However, if we write the equation as \(cq^{2} = -(ap^{2} + bpq)\) and see different cases by taking \(p, q\) to be even or odd, we see that the equation may be satisfied only when both \(p, q\) are even. But that contradicts the condition that \(p, q\) are coprime.This is what Colin says. – Siladitya Basu · 2 years, 11 months ago

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– Colin Tang · 2 years, 11 months ago

Ok that makes sense now ThanksLog in to reply