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?
Roots – Do Irrational Conjugates Always Come in Pairs?
graphing-functionsquadraticsroots
Best Answer
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.