wat abt the bracket is it allowed for this kind of operation?
subseven
2006-11-17 13:18:30 · 9 answers · asked by rEmo 1 in Science & Mathematics ➔ Mathematics
Yes. If you multiply it out it is the same. The square kills off that minus sign because any real number squared is positive.
z
2006-11-17 13:22:34 · answer #1 · answered by Rich Z 7 · 0⤊ 2⤋
Ok...
Look at the exact way this problem is presented :
Could be - ((b-c)^2) :
Answer: -(b^2 + c^2 - 2bc)
and if you distribute the negative inside the () you will get :
-b^2 - c^2 + 2bc
Now if it is ( -(b-c))^2 you will get
(c-b)^2 ---> (first distributing the negative in the (), then solving just like before....
I think you get the idea...
Good LUCK
2006-11-17 21:25:19 · answer #2 · answered by saman f 1 · 0⤊ 0⤋
no.
the - is not in ( ) so you perform ^2 first and then apply - afterwards.
-(b-c)^2 = - (b^2 + c^2 - 2bc)
or - b^2 - c^2 + 2bc
Shortcut to Check: plug in b=2 and c=1 into both expressions.
If your answer comes out different, the two are not equal,
such as -1 not equal to 1.
2006-11-17 21:25:47 · answer #3 · answered by emilynghiem 5 · 0⤊ 0⤋
No.
-(b-c)^2 = -(b^2+c^2-2bc) = -b^2 - c^2 + 2bc.
For the minus sign to get squared too, it would have had to be [-(b-c)]^2.
2006-11-17 21:23:08 · answer #4 · answered by Anonymous · 0⤊ 0⤋
no
think about in these terms
take the minus sign in -(b-c) to be replaced by (-1)
so now it would be (-1)(b-c)^2
i think you get the idea after that
2006-11-17 22:02:29 · answer #5 · answered by k soni 2 · 0⤊ 0⤋
Nope.Just expand out and it's different.
2006-11-17 21:47:53 · answer #6 · answered by A 150 Days Of Flood 4 · 0⤊ 0⤋
Yes.
2006-11-17 21:20:39 · answer #7 · answered by EternalBlueMemory 4 · 0⤊ 2⤋
no.
-(b-c)^2
-(b^2-2bc+c^2)
-b^2+2bc-c^2
2006-11-17 21:28:36 · answer #8 · answered by yupchagee 7 · 0⤊ 0⤋
no, the signs are wrong
2006-11-17 21:38:57 · answer #9 · answered by nobody 1 · 0⤊ 0⤋