algebra precalculus - Sum of roots rational but product irrational

Suppose that $x_1,x_2,x_3,x_4$ are the real roots of a polynomial with integer coefficients of degree $4$, and $x_1+x_2$ is rational while $x_1x_2$ is irrational. Is it necessary that $x_1+x_2=x_3+x_4$?

For example, the polynomial $x^4-8x^3+18x^2-8x-7$ has roots $$x_1=1-\sqrt{2},x_2=3+\sqrt{2},x_3=1+\sqrt{2},x_4=3-\sqrt{2}.$$
It holds that $x_1+x_2$ is rational while $x_1x_2$ is irrational, and we have $x_1+x_2=x_3+x_4$.

Answer

Suppose that all the roots are non zero. Put $x_1+x_2=u$, $a=x_1x_2$, $x_3+x_4=v$, $b=x_3x_4$, we suppose that $u\in\mathbb{Q}$, then it is also the case for $x_3+x_4$, as the sum of all the roots is rational, and that $a,b$ are irrationals (as the product $ab$ is rational, if $a$ is irrational, then it is also the case for $b$). The polynomial with roots $x_1,x_2,x_3,x_4$ is
$$P(x)=(x^2-ux+a)(x^2-vx+b)=x^4-(u+v)x^3+(a+uv+b)x^2-(av+bu)x+ab$$

By hypothesis, we get that $a+b+uv$, $av+bu$, and $ab$ are in $\mathbb{Q}$. Writing

$av+bu=(a+b)v-bv+bu$, we see that $b(u-v)$ is rational. As $b$ is not, this imply $u=v$.

If now $x_3x_4=0$, then it is not possible that $x_3=x_4=0$,(in this case $P(x)=x^4-ux^3+ax^2$) as $a\not \in \mathbb{Q}$, and if $x_4=0$ and $x_3\not = 0$, we get that $x_3\in \mathbb{Q}$, and $x_3\in \mathbb{Q}$ is not possible again, as $P(x)=(x-x_3)(x^2-ux+a)=x⁴+..-x_3ax$.