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Rational Quadratic

Problem

Given that a and p are positive integers and p is prime, find the values of a for which the following quadratic has rational roots.

ax2 minus px minus p = 0


Solution

Begin by dividing through by p: (a/p)x2 minus x minus 1 = 0.

To have rational roots, the discriminant, 4a/p + 1 = k2, where k = r/s and let us suppose that HCF(r,s) = 1.

therefore 4a/p = k2 minus 1 = r2/s2 minus 1 = (r2 minus s2)/s2, giving 4a = p(r2 minus s2)/s2

Clearly 4a is integer which can only happen if s = 1, as neither p nor r2 is divisible by s2 for any other value; that is, k is integer.

So 4a/p = k2 minus 1 = (k + 1)(k minus 1).

This leads to a = p(k + 1)(k minus 1)/4 implies k must be odd, as a is integer.

Let k = 2m + 1, so that a = p(2m + 2)(2m)/4 = pm(m + 1).

That is, for the quadratic to have rational roots, a must be a multiple of p, moreoever it must be of the form pm(m + 1).

Prove that the roots will be of the form 1/c and -1/(c + 1).

Problem ID: 247 (28 Oct 2005)     Difficulty: 3 Star

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