This is an exercise in Carlo Bourlet's textbook, LECONS D'ALGEBRE ELEMENTAIRE. Rationalising the following equation,
$$ \sqrt{A}+\sqrt{B}+\sqrt{C}+ \sqrt{D}+\sqrt{E}=0$$
I tried to square both sides of
$$ \sqrt{A}+\sqrt{B}+\sqrt{C}=-\sqrt{D}-\sqrt{E}$$
and I obtained,
$$ \frac{A+B+C-D-E}{2}=\sqrt{DE}-\sqrt{AB}-\sqrt{AC}-\sqrt{BC}$$
I tried to square both sides again, but I lost in too many square roots.
In Carlo Bourlet's book, as a example of three-square case, $$ \sqrt{A}+\sqrt{B}+\sqrt{C}=0$$ The rational equation can be obtained by two times squares, $$ A^2+B^2+C^2-2AB-2AC-2BC=0$$ This result can be obtained from Bourlet's construction,
$$\sqrt{A}+\sqrt{B}+\sqrt{C}=0$$
$$\sqrt{A}-\sqrt{B}-\sqrt{C}=0$$
$$\sqrt{A}-\sqrt{B}+\sqrt{C}=0$$
$$\sqrt{A}+\sqrt{B}-\sqrt{C}=0$$
Mutipling the above four equations we also obtain the rational form of the equation. I wonder for the five-square root case, can I do a similar thing?
To continue your initial "Shift terms around so that when you square, you reduce the number of radicals" approach:
Do exactly that again.
One possible way to reduce to 3 radicals (with coefficients) is:
To continue the "multiplying the equations":
Yes, and you can multiply out the 16 conjugate terms of the form $ \sqrt{A} \pm \sqrt{B} \pm \sqrt{C} \pm \sqrt{D} \pm \sqrt{E}$.
If $\sqrt{A}$ remains problematic, then multiply out the 32 conjugate terms where we also use $ \pm \sqrt{A}$.
Note: For $\geq 5$ terms, this is my recommended approach.