SOLUTION: Please help me solve this question x^2+y^2+z^2=2(x+z-1)then the value of x^2+y^2+z^2

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Question 1043596: Please help me solve this question x^2+y^2+z^2=2(x+z-1)then the value of x^2+y^2+z^2
Found 2 solutions by KMST, ikleyn:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2By%5E2%2Bz%5E2=2%28x%2Bz-1%29
x%5E2%2By%5E2%2Bz%5E2=2x%2B2z-2
x%5E2-2x%2By%5E2%2Bz%5E2-2z=-2
x%5E2-2x%2B1%2By%5E2%2Bz%5E2-2z%2B1=-2%2B1%2B1
%28x-1%29%5E2%2By%5E2%2B%28z-1%29%5E2=0
That describes the point with system%28x=1%2Cy=0%2Cz=1%29 ,
and substituting those values for the variables we get
x%5E2%2By%5E2%2Bz%5E2=1%5E2%2B0%5E2%2B1%5E2=highlight%282%29

Answer by ikleyn(52890) About Me  (Show Source):
You can put this solution on YOUR website!
.
Please help me solve this question x^2+y^2+z^2=2(x+z-1)then the value of x^2+y^2+z^2
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If  x%5E2%2By%5E2%2Bz%5E2 = 2%28x%2Bz-1%29,  then

x%5E2+-+2x+%2B+y%5E2+%2B+z%5E2+-+2z = -2,

%28x-1%29%5E2+-+1+%2B+y%5E2+%2B+%28z-1%29%5E2+-+1 = -2,

%28x-1%29%5E2+%2B+y%5E2+%2B+%28z-1%29%5E2 = 0.

We have the sum of squares of three real numbers equal to zero.
Hence, each of the three numbers is equal to zero:

x-1 = 0,  y = 0  and  z-1 = 0.

In other words, x = 1, y = 0, z = 1.

Now x%5E2+%2B+y%5E2+%2B+z%5E2%7D%7D+=+%7B%7B%7B1%5E2+%2B+0%5E2+%2B+1%5E2 = 2.

Solved.