SOLUTION: Hi! I was wondering: Does [B^2 + A^2] = [(A+B)^2]? Why or why not? I've tried proving this with numerous numbers, and it only works when either A or B is zero. On my homework,

Algebra ->  Distributive-associative-commutative-properties -> SOLUTION: Hi! I was wondering: Does [B^2 + A^2] = [(A+B)^2]? Why or why not? I've tried proving this with numerous numbers, and it only works when either A or B is zero. On my homework,      Log On


   

Question 279759: Hi! I was wondering:
Does [B^2 + A^2] = [(A+B)^2]? Why or why not? I've tried proving this with numerous numbers, and it only works when either A or B is zero. On my homework, I have to say why or why not this equation is true, and prove or disprove it with algebra. I don't get it?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
It is not true. So A%5E2%2BB%5E2%3C%3E%28A%2BB%29%5E2 in general. To show that it is false, simply find values of A and B which contradict the equation.

By the way, the only solutions to A%5E2%2BB%5E2=%28A%2BB%29%5E2 are when either A=0 or B=0. If A=0, then 0%5E2%2BB%5E2=%280%2BB%29%5E2 which then becomes B%5E2=B%5E2 which is true for all B. This argument works similarly when B=0. For any other values of A and B, the equation is false.


Here's why the only solutions are A=0 or B=0:


A%5E2%2BB%5E2=%28A%2BB%29%5E2 Start with the given equation.


A%5E2%2BB%5E2=A%5E2%2B2AB%2BB%5E2 FOIL


0=2AB Subtract A%5E2 from both sides. Subtract B%5E2 from both sides.


2AB=0 Rearrange the equation.


AB=0 Divide both sides by 2.


A=0 or B=0 Use the zero product property.