why is the square of the sum of two numbers is different from the sum of the squares of two numbers?

Answer 1

A more correct question would have been: can they be the same?

In maths, it is more common to have inequality than equality. That is why we spend so much time analysing equalities: they are rare beasts that provide us with powerful tools.

a^2 + b^2 ?=? (a+b)^2

a^2 + b^2 ?=? (a+b)*(a+b)

a^2 + b^2 ?=? a^2 + 2ab + b^2

(take away a^2 + b^2 from both sides)

0 ?=? 2ab

True only if a=0 or b=0

Otherwise, the difference is 2*a*b (a number not equal to 0).

Answer 2

(a + b)^2 = a^2 + 2ab + b^2 therefore unless a = 0 or b =0 the 2ab in the middle is going to be the difference.
If you mean why does the first formula work, then try drawing a square. Divide two adjacent sides in the same way unequally (if you see what I mean) and call the lengths a and b along each side
(a > b). This gives you a large square with area a^2, a small square with area b^2 and two rectangles each with area ab. These must add up to the area of the original square which is (a + b)^2.

Answer 3

u can use a simple working to find out why:

square of the sum of 2 numbers:
(1)------ (x+y)^2 = x^2 + 2xy + y^2

sum of the squares of two numbers:
(2)------ (x)^2 + (y)^2 = x^2 + y^2

the difference is for the first equation, u haf a 2xy, which u dun haf in the 2nd equation. so (1) is not the same as (2)

Answer 4

The operation of squaring (or cubing etc...) is not distributive over a plus or minus sign because then the order of operations would be violated.