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If $b$ is rational and $c$ is real, can $x^2 + bx + c$ have one rational root and one irrational root? Use only the definition of rational numbers to prove.

I think that it can't because since $b$ is rational, if two real roots exist then both are either rational or irrational, however, I'm not sure how to prove this using only the definition of rational numbers i.e. it is a number that can be expressed as $\frac{x}{y} $ where $x$ and $y$ are integers $y$ being nonzero. So you can't assume stuff like rational + irrational = irrational.

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  • $\begingroup$ By real root you mean irrational root? $\endgroup$ – R.V.N. Feb 17 at 17:07
  • $\begingroup$ And by "$c$ is real" do you mean "$c$ is irrational"? $\endgroup$ – Michael Cohen Feb 17 at 17:08
  • $\begingroup$ Welcome to Mathematics Stack Exchange. Note that the sum of roots is $-b$ $\endgroup$ – J. W. Tanner Feb 17 at 17:09
  • $\begingroup$ @R.V.N. yes my bad $\endgroup$ – TurboAverage419 Feb 17 at 17:09
  • $\begingroup$ @MichaelCohen no c is any real number $\endgroup$ – TurboAverage419 Feb 17 at 17:09
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Of course, note that all rational numbers are indeed real.


I suppose you wanted to ask "Is it possible that one root is rational and the other is not". The answer to that is then "no".

Note that the sum of the two roots is $-b \in \Bbb Q$. Using the definition of rationals, it follows that the difference of two rationals is rational. Conclude that $\text{rational} + \text{irrational} = \text{rational}$ is not possible.

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  • $\begingroup$ Is rational numbers being closed under add/sub enough to conclude rational + irrational = rational not possible? Since we can't assume anything it seems there is no proof that there does not exist an irrational number that can satisfy the above equation. $\endgroup$ – TurboAverage419 Feb 17 at 17:17
  • $\begingroup$ If $q_1, q_2$ are rationals and $r$ an irrational such that $q_1 + r = q_2$, what do you get by subtracting $q_1$ from both sides? $\endgroup$ – Aryaman Maithani Feb 17 at 17:20
  • $\begingroup$ Oh I see now, if rational numbers are closed under subtraction that implies that an irrational number cannot be the difference of two rational numbers? $\endgroup$ – TurboAverage419 Feb 17 at 17:23
  • $\begingroup$ Cannot be the difference of two rational numbers, yes $\endgroup$ – Aryaman Maithani Feb 17 at 17:23
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    $\begingroup$ Argh, another typo! Thank you! $\endgroup$ – TurboAverage419 Feb 17 at 17:25
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Hint

Suppose $r$ is a rational root and $j$ is a irrational root.

$$r+j=-b\to j=-(b+r).$$

Can you see the problem?

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    $\begingroup$ Thanks, your comment rephrased what someone else said and helped clarify it! $\endgroup$ – TurboAverage419 Feb 17 at 17:26

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