Question 154087
Algebraic Explanation:


{{{(a+b)^2=a^2 + b^2}}} Start with the given equation



{{{a^2+2ab+b^2=a^2 + b^2}}} FOIL the left side
 


{{{2ab=a^2 + b^2-a^2-b^2}}} Subtract {{{a^2}}} from both sides. Subtract {{{b^2}}} from both sides



{{{2ab=0}}} Combine like terms



Case 1: Solving for "a"


{{{a=0/(2b)}}} Divide both sides by {{{2b}}}



{{{a=0}}} Reduce


Or...


Case 2: Solving for "b"



{{{b=0/(2a)}}} Divide both sides by {{{2b}}}



{{{b=0}}} Reduce



So {{{(a+b)^2=a^2 + b^2}}} is true only if "a" or "b" is 0 (or both are zero). Otherwise, the equation is false. So in general {{{(a+b)^2 <> a^2 + b^2}}} 




Numerical Example:


{{{(a+b)^2=a^2 + b^2}}} Start with the given equation



{{{(5+4)^2=5^2+4^2}}} Plug in a=5 and b=4 (any two numbers will do as long as they are both not equal to zero)



{{{(9)^2=5^2+4^2}}} Add



{{{81=25+16}}} Square each term



{{{81=41}}} Add



Since {{{81=41}}} is <b>not</b> true, this means that we've just shown that {{{(a+b)^2=a^2 + b^2}}} is <b>not</b> true for all values of "a" and "b". 


So once again, in general {{{(a+b)^2 <> a^2 + b^2}}}