SOLUTION: Here is my question: If 2x = a+b+c, show that (x-a)^2 + (x-b)^2 + (x-c)^2 + x^2 = a^2 + b^2 + c^2. Thanks in advance! :)

Algebra ->  Expressions-with-variables -> SOLUTION: Here is my question: If 2x = a+b+c, show that (x-a)^2 + (x-b)^2 + (x-c)^2 + x^2 = a^2 + b^2 + c^2. Thanks in advance! :)      Log On


   

Question 1001892: Here is my question:
If 2x = a+b+c, show that (x-a)^2 + (x-b)^2 + (x-c)^2 + x^2 = a^2 + b^2 + c^2.
Thanks in advance! :)
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
%28x-a%29%5E2+%2B+%28x-b%29%5E2+%2B+%28x-c%29%5E2+%2B+x%5E2 ≟ a%5E2+%2B+b%5E2+%2B+c%5E2

Two expressions are equal if they differ by zero.  So the procedure 
we will use is to show that the difference between the two expressions 
above that we want to show equal, is zero.

So we examine this difference:



Remove the outer parentheses:

%28x-a%29%5E2+%2B+%28x-b%29%5E2+%2B+%28x-c%29%5E2+%2B+x%5E2-a%5E2+-+b%5E2+-+c%5E2%29

Arrange them so as to have three differences of squares:

%28x-a%29%5E2-a%5E2+%2B+%28x-b%29%5E2-b%5E2+%2B+%28x-c%29%5E2-c%5E2+%2Bx%5E2

Group the three differences of squares:



Factor the three differences of squares:



Remove the inner parentheses:

%28x-a-a%29%28x-a%2Ba%29+%2B++%28x-b-b%29%28x-b%2Bb%29+%2B+%28x-c-c%29%28x-c%2Bc%29%2Bx%5E2

Collect like terms:

%28x-2a%29%28x%29+%2B++%28x-2b%29%28x%29+%2B+%28x-2c%29%28x%29%2Bx%5E2

Distribute to remove the parentheses:

x%5E2-2ax+%2B++x%5E2-2bx+%2B+x%5E2-2cx%2Bx%5E2

Combine like terms:

4x%5E2-2ax+-2bx+-2cx

Factor out 2x

2x%282x-a-b-c%29

Substitute a+b+c for 2x since that is given:

%28a%2Bb%2Bc%29%28a%2Bb%2Bc-a-b-c%29

The second parenthetical expression becomes 0:

%28a%2Bb%2Bc%29%280%29

0

The two original expressions differ by 0 so they are equal.

Edwin