x2+6y=-17 ..................... (1)
y2+4z=1 ..................... (2)
z2+2x=2 ...................... (3)
now add 1,2&3
we get x2+6y+y2+4z+z2+2x=-17+1+2
=-14
adding 1,4,9 on LHS and RHS
i.e.
x^2 +6y+y^2+4z+z^2+2x++9+4+1= -14+1+9+4
therefore
(x^2+2x+1)+(y^2+6y+9)+(z^2+4z+4)=0
(x+1)^2+(y+3)^2+(z+2)^2=0
we did the above step by adding (1^2=1)+(2^2=4)+(3^2=9) on both side of the equation and factorizing the LHS.
so the equation becomes (x+1)^2+(y+3)^2+(z+2)^2=0
therefore to get 0 each of the three squares should be equal to 0
(x+1)^2=0 x=-1
(y+3)^2=0 y=-3
(z+2)^2=0 z=-2
so the value of x2+y2+z2=(-1)^2+(-3)^2+(-2)^2
=1+4+9=14
ans-14
#guest4
yes i have checked the solution by using substituition method. and if still you are not satisfied by my solution then plz be clear with your question so that even i can be more clear in my explaination..!!!