Total Answers: 8 [Go Back]
um in one equation the -x is surrounded by brackets the other is not :P nah sorry i'm just being a smart ass, i really don't know :S perhaps ask your maths teacher?
Sun, 01 Aug 2010 09:11:31 GMT
they are the same
Sun, 01 Aug 2010 09:11:44 GMT
(-x)^2 is always positive, since that's -x multiplied by -x -x^2 is always negative, since it's x multiplied by x, then negative same with -(x)^y~ -x^y is the same as -(x)^y this is because of BEMDAS, just not written, but powers (that is, ^2) takes precedence over multiplication/addition and the rest
Sun, 01 Aug 2010 09:13:36 GMT
(- x)^2 (-x)*(-x) +x^2 [the bracket means whole square, the power implies to also] -x^2 -(x*x) -x^2 [it means only x^2 excluding the (-)] in other way, (-x)^2 [(-1)*(x)]^2 (-1)^2 (x)^2 1 x^2 x^2 - x^2 (-1) (x)^2 -x^2 You see -(x)^y (x)^y always it will be true whether y is even or odd. -(x)^y and -x^y are same.
(-x)^y x^y when y=even integer (-x)^y -x^y when y=odd integer. I think its clear now.
(-x)^y x^y when y=even integer (-x)^y -x^y when y=odd integer. I think its clear now.
Sun, 01 Aug 2010 09:25:34 GMT
Recall that (a b)^2 a^2 b^2 Then (-x)^y (-1)^y x^y Here we see that for even y, (-x)^y x^y and for odd y, (-x)^y -x^y If you are not sure about the difference between -x^2 and (-x)^2, you could just expand the terms.
that x^n x x x x and there are n terms.
-x^2 (x x) -x^2 and (-x)^2 (-x) (-x) (-1)^2 x^2 x^2
that x^n x x x x and there are n terms.
-x^2 (x x) -x^2 and (-x)^2 (-x) (-x) (-1)^2 x^2 x^2
Sun, 01 Aug 2010 09:29:01 GMT
(-x)^2 The square is applied to x as well as the negative sign to give you x^2 -x^2 The square is only applied to the x term -(x)^y and -x^y are essentially the same because no matter what number y is, the negative term will always be present as the it is not factored into the power. Also, It's pointless to brackets to the x term. (-x)^y because the brackets include the negative sign, the final term will be either positive if y is an even number or negative if y is an odd number.
Sun, 01 Aug 2010 09:32:44 GMT
there is none bc both are ^2 if i was ^3 or any other odd then one would be and the other would be and with the ^y you would need to find find before knowing
Sun, 01 Aug 2010 09:39:21 GMT
It's quite simple, remember the exponent means repeated multiplication of the base, so: (-x)^2 (-x)(-x) x^2 While in the other one: -x^2 (-1)(x)(x) -x^2 They are not equal because, after we evaluate each one, x^2 is not equal to -x^2. Once you remember the order of operations this becomes evident and you don't need to factor like I did.
Sun, 01 Aug 2010 13:28:32 GMT
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