Solving a system of equations, three variables x,y,z here's a problem from this PRMO which I found online:
Three real numbers $x,y,z$ are such that $x^{2}+6y=-17$, $y^{2}+4z=1$, $z^{2}+2x=2$. What is the value of $x^{2}+y^{2}+z^{2}$?
Ok I am quite confused about what to do because simple manipulations like adding the three equations don't seem to work (or at least I don't see anything yet). It would be really helpful if someone could tell me the approach or point out what I'm missing. Would be great if you could just give a small hint first instead of the whole answer immediately. Thanks!
 A: Hint: From the first equation we get
$$y=-\frac{1}{6}\left(17+x^2)\right)$$ plugging this in the second equation and solving this for $z$
$$z=\frac{1}{4}\left(1-\frac{1}{36}\left(17+x^2\right)^2\right)$$ so we get
$$\frac{1}{16}\left(1-\frac{1}{36}(17+x^2)^2\right)^2+2x=2$$
Factorizing we obtain
$$\frac{(x+1)^2 \left(x^2+4 x+31\right) \left(x^4-6
   x^3+64 x^2-210 x+727\right)}{20736}=0$$
A: Adding all equations we have
$$
x^2+2x+\alpha+y^2+6y+\beta+z^2+4z+\gamma = 0\\
\alpha+\beta+\gamma = 14
$$
Now for 
$$
x^2 +2x+\alpha = 0\\
y^2+6y+\beta=0\\
z^2+4z+\gamma = 0
$$
making the discriminants
$$
2^2-4\alpha = 0\to \alpha = 1\\
6^2-4\beta = 0\to\beta = 9\\
4^2-4\gamma = 0\to\gamma = 4
$$
then the sum gives
$$
\left(x+\frac 22\right)^2+\left(y+\frac 62\right)^2+\left(z+\frac 42\right)^2=0
$$
so the solution is $x = -1, y = -3, z = -2$
and then $x^2+y^2+z^2 = 14$
A: Adding all three can work actually ,
x² + 6y = -17,
y² + 4z = 1
z² + 2x  = 2
x² + y² + z²  + 2x + 4z  + 6y  = -14 , and if we want to make rhs zero , we can do
(x + 1)² - 1  + (y + 3)² -9  + (z + 2)² - 4  = -14 which is
(x + 1)²  + (y + 3)²   + (z + 2)²   = 0
keeping each term zero
we find that x = -1 , y = -3 , z = -2
x² + y² + z²  = 14
