What is (s/2)^2 and what is (1/2)^2? See my ADDED details please?

Question

Explain how do you get the answer and also what is the square root of 4a^2-4ab+b^2+c^2. Also please tell me the square root of b^2+2ab+a^2+c^2.Thank YOU and be sure that if you answer all three of my question you will get picked as best answer.

Answer 1

(s/2)^2 = s^2 / 4
(1/2)^2 = 1/4

sqrt( 4a^2-4ab+b^2+c^2) = sqrt ( (2a-b)^2 + c^2) but you cannot simplifly further.

sqrt (b^2+2ab+a^2+c^2) = sqrt ( (b+a)^2 + c^2), but once again, you cannot simply anymore .

the person who answer above, is wrong...
never ever is sqrt (a^2+b^2) = a+b, never .
(example: sqrt ( 3^2 + 4^2) = sqrt (9+16) = sqrt (25) = 5
but 3 +4 = 12).....

Answer 2

(s/2)^2 is simply s^2 divided by 2^2 = (s^2)/4

(1/2)^2 is done the same way. Square the top and square the bottom. You get 1/4.



sqrt = square root
sqrt of 4a²-4ab+b² +c²

spilt them up first:
4a² -4ab+b² is obviously (2a+b)² . Remember the (a+b)² is equal to (a + b)² = a² + 2ab + b²
take the sqrt of that and you get 2a+b
next do sqrt of c² , which is c.
so the answer for that is 2a+b+c




sqrt of b²+2ab+a²+c².
same thing. b²+2ab+a² = (a+b)². sqrt of that is a+b.
sqrt of c² is c
answer: a+b+c