Do Irrational Conjugates always come in pairs?

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.

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.