Show that if a quadratic has rational coefficients and $x + y \sqrt{z}$ is a root, then $x - y \sqrt{z}$ must also be a root.

Given $$x, y$$ and $$z$$ are rational and $$\sqrt{z}$$ is not, show that if $$x + y\sqrt{z}$$ is a root of a quadratic with rational coefficients, then $$x - y\sqrt{z}$$ is also a root.

In the situation the discriminant of the equation is positive. I think I have an idea how to start but haven't been able to go any further.

$$$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = x \pm y \sqrt{z}$$$$

Is my setup incorrect? How can I approach this problem?

Maybe try using the following relationships of the roots and coefficients of a quadratic polynomial: $$\text{Let } a, b, c \in \mathbb{Q}.$$
$$ax^2 + bx + c = a(x-x_1)(x-x_2) = a(x^2 -(x_1 + x_2) + x_1 \cdot x_2) = ax^2 - a(x_1 + x_2)x +a \cdot x_1 \cdot x_2.$$ From here we get the following system of equalities: $$\begin{cases} x_1 + x_2 = -\frac{b}{a} \\\ x_1 \cdot x_2 = \frac{c}{a} \end{cases}$$
Notice that if $$x_1 = x + \sqrt{z}$$, then $$x_1 + x_2 = x + \sqrt{z} + x_2 = -\frac{b}{a} \in \mathbb{Q}$$, then we can solve for $$x_2$$: $$x_2 = (-\frac{b}{a} - x) - \sqrt{z} = m - \sqrt{z}, \text{ where } m \in \mathbb{Q}$$ Now let's use the second relationship we proved: $$x_1 \cdot x_2 = (x+\sqrt{z})(m-\sqrt{z}) = (xm - z) + (m - x)\sqrt{z} = \frac{c}{a} \in \mathbb{Q}$$ If $$m - x \neq 0$$ we could solve for $$\sqrt{z}$$ to get that it must be rational, but this contradicts our assumption that it is irrational, hence $$m = x$$, which implies $$x_2 = m -\sqrt{z} = x - \sqrt{z}$$.