Discussion
−24 − (x2 − 1)2 =
| (A) | − x4 + 2x2 +15 |
| (B) | ... |
| (C) | ... |
| (D) | ... |
| (E) | ... |
| (F) | ... |
The solution is
Posted: 01/15/2012 12:23
Привет! Который палец онлайн для форуме?
Posted: 01/15/2012 14:16
Matt, we'll start with:
-2^4 - ( x^2 - 1 )^2
Now we'll use the distributive property (a-b)^2 = a^2 -2ab + b^2
-2^4 - ( (x^2)^2 - 2x^2 - 1^2 )
-2^4 - ( x^4 - 2x^2 - 1^2 )
-2^4 -x^4 + 2x^2 + 1^2
-x^4 + 2x^2 + 1 - 16
-x^4 + 2x^2 - 15
Niels
-2^4 - ( x^2 - 1 )^2
Now we'll use the distributive property (a-b)^2 = a^2 -2ab + b^2
-2^4 - ( (x^2)^2 - 2x^2 - 1^2 )
-2^4 - ( x^4 - 2x^2 - 1^2 )
-2^4 -x^4 + 2x^2 + 1^2
-x^4 + 2x^2 + 1 - 16
-x^4 + 2x^2 - 15
Niels
Posted: 01/16/2012 16:17
Where does the 2x^2 come from?
Posted: 01/16/2012 19:41
Like I said earlier, using the (special case) distributive property, I will show it with FOIL.
(a-b)^2
(a-b)(a-b)
a^2 -ab -ba +b^2
a^2 -ab -ab +b^2
a^2 -2ab +b^2
If you plug in (x^2) for a, and (1) for b, it will get you the -2ab >> -2(x^2)(1) ergo -2x^2
Niels
(a-b)^2
(a-b)(a-b)
a^2 -ab -ba +b^2
a^2 -ab -ab +b^2
a^2 -2ab +b^2
If you plug in (x^2) for a, and (1) for b, it will get you the -2ab >> -2(x^2)(1) ergo -2x^2
Niels
Posted: 06/20/2012 02:04
I have been studing this problem for about a few mins now hoping to understand where (a) and (b) came from.. I sure you know way more then u do on this topic.. Plz explain?
Posted: 06/20/2012 02:06
Never mind I understand sorry
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Edit
Posted: 01/29/2012 18:25
Where does the -2x^2 came from, and why they changed -1 to positive 1?
Posted: 01/30/2012 00:47
Serigne, I guess you didn't look at this forum thread first.
The sign changes done are due the priority rules, parentheses first and when there's a negative sign in front, we distribute the sign over all terms in the parentheses by multiplying each term with -1
Example A, with a positive sign in front of the parentheses:
8 + ( 5 - 3 + 2 )
8 + (1)( 5 - 3 + 2 )
8 + 5 - 3 + 2
= 12
Example B, with a negative sign in front of the parentheses:
8 - ( 5 - 3 + 2 )
8 + (-1)( 5 - 3 + 2 )
8 - 5 + 3 - 2
= 4
Niels
The sign changes done are due the priority rules, parentheses first and when there's a negative sign in front, we distribute the sign over all terms in the parentheses by multiplying each term with -1
Example A, with a positive sign in front of the parentheses:
8 + ( 5 - 3 + 2 )
8 + (1)( 5 - 3 + 2 )
8 + 5 - 3 + 2
= 12
Example B, with a negative sign in front of the parentheses:
8 - ( 5 - 3 + 2 )
8 + (-1)( 5 - 3 + 2 )
8 - 5 + 3 - 2
= 4
Niels
Posted: 02/29/2012 17:46
Im not really good in math but im taking the sat ... In march 10th in im really lost
Posted: 10/28/2012 22:16
-2^4 = +16
Not -16
Not -16
Posted: 10/28/2012 22:19
The answer should be A
Posted: 12/04/2012 16:02
Sorry Katie, I think there's something unusual on your answer. It should be (C) instead of (A). I think I know what you did wrong. You have elevated -2 to the fourth power instead of elevating 2 to the fourth power and putting the negative sign. Am I right?
Posted: 03/07/2013 04:51
Yes, exactly.
Posted: 10/28/2012 22:20
Oh never mind
Posted: 11/03/2012 12:51
I don't get where the 2x^2 come from
Posted: 11/03/2012 14:46
Kary, that's just from "foiling" the terms inside the parentheses that are raised to the power of 2, which means they are multiplied by themselves.
Posted: 11/03/2012 14:50
Kary,
(x^2 - 1)^2
=(x^2 - 1) (x^2 - 1)
= x^4 - 2x^2 +1
Hope that helps
(x^2 - 1)^2
=(x^2 - 1) (x^2 - 1)
= x^4 - 2x^2 +1
Hope that helps
Posted: 11/03/2012 16:00
Thanks now I understand
Posted: 03/07/2013 04:48
(-2)^4=16, not -16.
-2*-2*-2*-2=16.
-2*-2*-2*-2=16.
Posted: 05/20/2013 16:29
Mantek, you are right. But that's not what the problem is asking. Not (-2)^4, but -2^4 ....
Posted: 05/19/2013 18:04
Where does the (-1) come from ? 2 doesn't apply as a variable like "x" in an equation so why is there a 1 there ?
Posted: 05/20/2013 16:30
Andres Tirado, the explanation is just showing that a negative sign is the same operation as multiplying a number or variable with -1.
Posted: 05/30/2013 17:33
Are choice B and E the same answer or is there something I'm missing ?
Posted: 06/06/2013 22:41
Andres, yes there is something you are missing. Please look more closely.
Posted: 06/11/2013 21:37
I see it, thanks