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If the coefficients of a quadratic equation \( ax^2 + bx + c = 0 (a \neq 0 ) \) , are all odd integers, show that the roots cannot be rational.

Suppose m and n are the roots. Then \( \frac{c}{a} = mn \).

Hence if one of them is rational then the other one is rational as well.

If possible say both roots are rational. Then

\( m = \frac{p}{q} , n = \frac {r}{s} \) where p,q,r,s are integers and g.c.d. (p,q) = 1, g.c.d.(r, s) = 1

Also \( mn = \frac{p}{q} \times \frac {r}{s} = \frac {c}{a} \).

Note that both r, s cannot be even (as their g.c.d. is 1). If possible say s is even, then r is odd. Also if s is even, p has to be even (since the 2's in s must be canceled out, as the product of the two fractions ultimately is \( \frac {c}{a} \) which has odd in the numerator and odd in the denominator). Since p is even, therefore, q must be odd.

So here are the cases:

Case 1: p is even (say = 2p', q is odd, r is odd, s is even (say = 2s') (and other similar cases where crosswise numbers are even and other pair being odd)

Then \( m+n = \frac{p}{q} + \frac {r}{s} = -\frac {b}{a} \).

This implies \( m+n = \frac{2p'}{q} + \frac {r}{2s'} = -\frac {b}{a} \).

Or \( m+n = \frac{4s'p' + qr}{2s'q} = -\frac {b}{a} \).

Since qr is odd and 4s'p' is even their sum is odd. But the denominator (2s'p) is even. Hence the numerator cannot cancel the 2's in the denominator. But that is a contradiction as this fraction is supposed to be equal to \( -\frac {b}{a} \) where denominator is odd.

Case 2: all of p,q,r,s are odd

\( m+n = \frac{p}{q} + \frac {r}{s} = -\frac {b}{a} \).

Here the numerator is ps+qr which is even (since p, q, r, s are odd). But the denominator qs is odd. Hence 2's in the numerator won't cancel out. Hence contradiction.

Thus the roots cannot be rational.

**What is this topic:**Theory of Equations**What are some of the associated concepts:**Sum and product of roots, Vieta's formula**Book Suggestions:**Theory of Equations by Burnside and Panton

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