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Okay - so, I have the expression $5x^2+2x-3$ and have found that odd integers ($2x+1$) output an even number ($2\times integer$). But, how do I know this is the LARGEST set of integers that outputs an even number?

Math1
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  • You could use the fact that when $x$ is even, $5x^2+2x-3$ is ALWAYS odd. Combine that with the fact that every integer is either even or odd, and it should work out. – 2012ssohn Nov 12 '16 at 02:03

2 Answers2

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When $x$ is even, $5x^2$ is even, $2x$ is even, and $3$ is odd, which means $5x^2 + 2x - 3$ is odd. When $x$ is odd, $5x^2$ is odd, $2x$ is even, and $3$ is odd, which means $5x^2 + 2x - 3$ is even. These are the only two possible cases for $x$.

Thus, for the expression to be even, $x$ must be odd, so the odd numbers are the largest set of integers such that $5x^2 + 2x - 3$ is even.

shardulc
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  • Thank-you, I had arrived at the solution but couldn't justify how I did. I just knew it to be true. This makes sense. – Math1 Nov 12 '16 at 02:23
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    @Math1 Thanks, you can upvote if you like the answer and accept it by clicking the green arrow – shardulc Nov 12 '16 at 02:32
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Write it as $n = 5x(x+1) - 3(x+1)$. The first term is even since the product of consecutive integers $x(x+1)$ is always even. Then $n$ will have the same parity as the second term which, given that $3$ is odd, is the parity of $x+1$. So in the end $n$ even $\iff x+1$ even $\iff x$ odd.

dxiv
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