Solve $2x^2+y^2-z=2\sqrt{4x+8y-z}-19$

Question

I am trying to solve the following equation. $$ 2x^2+y^2-z=2\sqrt{4x+8y-z}-19 $$ To get rid of the square root, I tried squaring both sides which lead to $$ (2x^2+y^2-z+19)^2=16x+32y-4z $$ which was too complex to deal with.

Also, I have tried some substitutions to simplify the equation, but none of them were working.

I believe that the equation could be solved with a appropriate substitution and factorization, yet I have no idea what to do.

Any hint or help is appreciated.

Answer 1

It's $$2x^2+y^2-4x-8y+18+4x+8y-z-2\sqrt{4x+8y-z}+1=0$$ 0r $$2(x-1)^2+(y-4)^2+(\sqrt{4x+8y-z}-1)^2=0,$$ which gives $$x-1=y-4=\sqrt{4x+8y-z}-1=0.$$ Can you end it now?

Answer 2

To make it a little easier to get Michael Rozenberg's answer, you can replace $4x+8y-z=t$: $$2x^2+y^2-(4x+8y-t)=2\sqrt{t}-19 \Rightarrow \\ 2(x^2-2x+1)+(y^2-8y+16)+(t-2\sqrt{t}+1)=0 \Rightarrow \\ 2(x-1)^2+(y-4)^2+(\sqrt{t}-1)^2=0 \Rightarrow \\ x-1=y-4=\sqrt{t}-1=0.$$