Obel1x wrote: $\star$ Suppose that the roots $p,q$ of the equation $x^2-x+c=0$ where $c \in \mathbb{R}$, are rational numbers. Prove that the roots of the equation $x^2+px-q=0$ are also rational numbers. The equation $x^2-x+c=0$ has 2 rational roots $p$ and $q$ Which gives $1-4c=t^2$ in which $t$ is a rational number. Then, we can suppose that $p=\frac{1-t}{2}$ and $q=\frac{1+t}{2}$ Hence, the discrimination of the second equation is $\Delta= p^2+4q=\left(\frac{t+3}{2}\right)^2$ $\Rightarrow \Delta$ is a rational number. In addition to $p$, $q$ are also rational numbers. So the equation $x^2-px+q=0$ also has 2 rational roots.

Obel1x wrote: $\star$ Suppose that the roots $p,q$ of the equation $x^2-x+c=0$ where $c \in \mathbb{R}$, are rational numbers. Prove that the roots of the equation $x^2+px-q=0$ are also rational numbers. $p+q=1$ and so the roots of $x^2+px-q$ are $-1$ and $q$, and both $\in\mathbb Q$