Prove the following proposition: If a, b and c are integers such that a is odd,

Question

Prove the following proposition:
If a, b and c are integers such that a is odd, b and c are even, but c is not divisible by 4, then the equation ax^2 + bx + c = 0 has no rational solutions.

Assume that every nonzero rational number may be expressed as the quotient of two integers that have no common factors
greater than 1. (state your proof method: direct proof, by contrapositive, or by contradiction) please explain all proof steps in detail and give all definitions of even, odd, and divisible by 4 needed for this question.

Explanation / Answer

the solutions to the equation are given by the well known formula ( -b + sqrt (b^2 - 4ac))/2a, and ( -b - sqrt (b^2 - 4ac))/2a for those to be rational we need that sqrt (b^2 - 4ac) to be and integer and for a root square of an integer to be an integer we need b^2 - 4ac to be a perfect square, that is that there is an integer m such that b^2 - 4ac = m^2 As b is even so is b^2 - 4ac, then m is even furthermore we have b^2 - m^2 = 4ac thus 4ac = (b + m)(b-m), as b and m are both even then we can write b = 2k m= 2n, sub this into the equality that we have 4ac = 4 (k + n)(k-n) thus ac = (k+n)(k-n), now k+n and k-n are both even or both odd, as c is even then k+n and k-n are both even write k+n= 2q and k-n= 2p, then ac= 4pq, therefore 4 divides ac, but a is odd then 4 divides c which is a contradiction. thus in this case b^2 - 4ac is not a perfect square and so sqrt(b^2-4ac) is irrational and the solutions to the equation are irrational.

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