Solve 3-Variable Equation with Real Solutions | √(z-y^2-6x-26)+x^2+6y+z-8=0

[anemone]

Find the real solutions for $\sqrt{z-y^2-6x-26}+x^2+6y+z-8=0$.

 

[WWGD]

One way to go about it is by moving some terms around to get rid of the square root.

 

[bob012345]

It's actually better to keep the square root.

 

[bob012345]

Find the real solutions for $\sqrt{z-y^2-6x-26}+x^2+6y+z-8=0$.

Here is my solution;

Spoiler

Given $$\sqrt{z-y^2-6x-26}+x^2+6y+z-8=0$$ we can write
$$z = 8 - x^2 -6y - \sqrt{z-y^2-6x-26}$$

we see that for real solutions $$z-y^2-6x-26 \ge 0$$ or $$z \ge y^2+6x+26$$

thus $$z = 8 - x^2 -6y - \sqrt{z-y^2-6x-26} \ge y^2+6x+26$$ or

$$- \sqrt{z-y^2-6x-26} \ge y^2+6x+26 +x^2 + 6y -8$$ the RHS can be condensed to

$$x^2 + y^2 + 6(x+y) +18 = 0$$ which can be written as

$$(x+3)^2 + (y+3)^2 =0$$ finally giving

$$- \sqrt{z-y^2-6x-26} \ge (x+3)^2 + (y+3)^2$$

This has only one real root $(x,y)= (-3,-3)$ when the LHS is zero then solving for $z$ gives $z=17$
so $(x,y,z) →(-3,-3,17)$
[\spoiler]