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Please forgive me for any mistakes while asking this question. While applying the quadratic formula to find out the roots, or using any other methods to find the roots, I observe that the roots of the equation $y=x^2-3$ are $\sqrt{3}$ and $-\sqrt{3}$. These roots evaluate to be irrational, and the roots always happen to come in irrational conjugates. If I do this with any other polynomial equation, the irrational solutions almost always come in conjugates. However, I have found a few exceptions to this, and I do not know why this occurs. Take the equation $-x^3-2x^2+2$, where two of the solutions are complex solutions. After applying long division, why do I get the other answer to be an irrational solution? I thought that if there is one irrational solution, surely there must be another, but this is not the case in the equation provided. Why is this so?

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    $\begingroup$ It's complicated. The subject you want is called Galois theory. $\endgroup$ – Qiaochu Yuan Jan 16 '18 at 23:34
  • $\begingroup$ I think I don't really understand the question. What is it exactly that you are wondering? You can produce a polynomial with any number of irrational roots you want, just choose whatever roots you want in $\mathbb{C}$ and then multiply the corresponding linear factors $\endgroup$ – Pedro Jan 16 '18 at 23:37
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    $\begingroup$ $x^3 + 3$ has one real irrational root, and two complex (also irrational) roots. However, complex roots (of polynomials with real or integer coefficients) always come in conjugate pairs. $\endgroup$ – Doug M Jan 16 '18 at 23:39
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    $\begingroup$ I think the motivation is that if $p + \sqrt{q}$ is a root of a quadratic (with rational coefficients) then $p - \sqrt{q}$ is as well (in many "elementary" math classes, expressions of the form $a + b$ and $a - b$ are called conjugates, no matter what type of things $a$ and $b$ are). [...] $\endgroup$ – pjs36 Jan 17 '18 at 0:13
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    $\begingroup$ [...] A thought on why this doesn't generalize neatly: conjugates (in the above sense) are strongly motivated by quadratics, in particular by the difference of squares identity $(a + b)(a - b) = a^2 - b^2$. But if $a$ and $b$ are cube roots, instead of square roots, then $a^2 - b^2$ is probably not rational; e.g., $(1 - \sqrt[3]{5})(1 + \sqrt[3]{5}) = 1 - \sqrt[3]{25}$. Conjugates (in the above sense) lose their niceness, if $a$ and $b$ aren't the "right" type of thing. But I don't really know. $\endgroup$ – pjs36 Jan 17 '18 at 0:13
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The product of all the roots of the polynomial is exactly the constant term of the polynomial, up to a factor of $\pm 1$. So if your polynomial has all integer coefficients and at least one root not in $\mathbb{Q}$, then it has to have at least one other root not in $\mathbb{Q}$ - otherwise, we would multiply all the roots and have an irrational number equal to a rational one.

These roots can, of course, be complex; in this case, either the real part or the imaginary part must be irrational. This is exactly what you're observing in this case. There is one real, irrational root and a pair of complex (conjugate) roots that have irrational real and imaginary parts.

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I think you are confusing irrational solutions with complex solutions.

Irrational solutions need not come in pairs. The equation $$ x^3 - 2 = 0 $$ has three roots. One is the irrational real number $\alpha = 2^{1/3}$.The other two are the complex conjugate pair $$ \alpha \left(\frac{-1 \pm i \sqrt{3}}{2} \right). $$

The complex roots of a polynomial with real coefficients always come in conjugate pairs.

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  • $\begingroup$ Please provide the explanation to why the irrational solutions do not need to come in pairs, otherwise just give me the link to where you found the explanation, as I am currently arguing with my teacher about this. $\endgroup$ – Yash Jain Jan 16 '18 at 23:54
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    $\begingroup$ I can't provide a better explanation for why the pairing of irrational real roots doesn't happen other than a counterexample, like the one in the answer. If you edit the question to provide an argument for why you or your teacher think real irrational roots should come in pairs then someone here may be able to point out the error. $\endgroup$ – Ethan Bolker Jan 17 '18 at 0:06
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Here is a real polynomial with two roots: one rational and one irrational: $$ P(x) = (x+3)(x+\sqrt{3}) $$ They are not irrational conjugates, naturally.

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