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

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?


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

Tags:

Algebra Precalculus

Systems Of Equations

Proof Writing

Sum Of Squares Method

Algebraic Equations

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