Can the sum of irrational square roots of two different rational numbers be another irrational square root of a rational number? Prove or refute that $\exists x, y, z \in$ with   $\mathbb {Q}$ and $x\neq y$ such that $\sqrt{x}+\sqrt{y}=\sqrt{z}$ and $\sqrt{x}, \sqrt{y}, \sqrt{z}\in \mathbb{R}\setminus\mathbb{Q}.$ 
If $\sqrt{x}, \sqrt{y}, \sqrt{z}\in \mathbb{R}$, we find that $\sqrt{1}+\sqrt{4}=\sqrt{9}$ or even $\sqrt{0}+\sqrt{1}=\sqrt{1}$ holds. Thus we are interested in the case where the roots are irrational numbers but the radicands are rational numbers. 
After squaring both members of the equality we get $x+2\sqrt{xy}+y=z\implies\sqrt{xy}=\frac{z-x-y}{2}$. This means that the product of $\sqrt{x}$ by $\sqrt{y}$ is a rational number. We can only conclude that both $\sqrt{x}$ and $\sqrt{y}$ are rational or irrational.  
I can't find any good aproach to this problem and I don't even know if this result is relevant. Thank you for advance.

Letting $x = 2$, $y = 8$, and $z=18$ works:
$$\sqrt 2 + \sqrt 8 = \sqrt 2 + 2 \sqrt 2 = 3 \sqrt 2 = \sqrt 18.$$
More generally, for any $a, b, c \in \mathbb Q$ we have
$$ \sqrt{ab^2} + \sqrt{ac^2} = b\sqrt a + c \sqrt a = \sqrt{a(b+c)^2}.$$
This is an example whenever $b \ne \pm c$ and $\sqrt a$ is irrational.
