simplify the expression
( 3^x +3^-x)^2 - (3^x - 3^-x)^2
the answer is 4 but i need to know the steps of how to do it so please show detail
2007-07-03 18:01:27 · 10 answers · asked by fhsdjfsd 2 in Science & Mathematics ➔ Mathematics
THE MOST SIMPLE SOLUTION:
It is difference of squares!
DO NOT EXPAND!
take a=3^x+3^-x, b=3^x-3^-x
in
a^2-b^2 = (a-b) (a+b)
we have
a-b = 2 * 3^-x
a+b = 2 * 3^x
( 3^x +3^-x)^2 - (3^x - 3^-x)^2
= (3^x +3^-x -3^x +3^-x) (3^x +3^-x +3^x -3^-x)
= 2 * 3^-x * 2 * 3^x
= 4
Because 3^x * 3^-x = 1
2007-07-03 18:52:57 · answer #1 · answered by Payam 2 · 0⤊ 1⤋
(3^x + 3^-x)^2 - (3^x - 3^-x)^2
= (3^x + 3^-x)(3^x + 3^-x) - (3^x - 3^-x)(3^x - 3^-x)
= (3^(2x) + 3^0 + 3^0 +3^(-2x)) - (3^(2x) - 3^0 - 3^0 + 3^(-2x))
= (3^2)^x + 1 + 1 + (3^-2)^x - (3^2)^x + 1 + 1 - (3^-2)^-x
= 9^x + 1 + 1 + (1 / 9)^x - 9^x + 1 + 1 - (1 / 9)^x
= 4
2007-07-03 18:37:10 · answer #2 · answered by mathjoe 3 · 1⤊ 0⤋
similar to mathjoe's approach....
(3^x + 3^-x)^2 - (3^x - 3^-x)^2
expanding gives...
= (3^x + 3^-x)(3^x + 3^-x) - (3^x - 3^-x)(3^x - 3^-x)
multiplying out gives...
= (3^(2x) + 3^0 + 3^0 +3^(-2x)) - (3^(2x) - 3^0 - 3^0 + 3^(-2x))
rearranging gives...
= 3^(2x) - 3^(2x) +3^(-2x) - 3^(-2x) +1+1+1+1
= 0 + 0 + 1+1+1+1 = 4
2007-07-03 18:57:18 · answer #3 · answered by Dr W 7 · 0⤊ 1⤋
( 3^x +3^-x)^2 - (3^x - 3^-x)^2
= (3^2x +2*3^0+3^-2x) - (3^2x -2*3^0 + 3^-2x)
= 3^2x +2 +3^-2x -3^2x +2 - 3^-2x
= 4
2007-07-03 19:00:27 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋
(3^x + 3^-x)^2 - (3^x - 3^-x)^2
[3^x + 1/(3^x)]^2 - [3^x - 1/(3^x)]^2
[3^x + 1/(3^x)][3^x + 1/(3^x)] - [3^x - 1/(3^x)][3^x - 1/(3^x)]
[(3^x^2) + (3^x/3^x) + (3^x/3^x) + 1/(3^x^2)] - [(3^x^2) - (3^x/3^x) - (3^x/3^x) + 1/(3^x^2)]
[(3^x^2) +1 +1 + 1/(3^x^2)] - [(3^x^2) -1 -1 + 1/(3^x^2)]
[(3^x^2) +2 +1/(3^x^2)] - [(3^x^2) - 2 +1/(3^x^2)]
(3^x^2) +2 +1/(3^x^2) - (3^x^2) + 2 - 1/(3^x^2)]
2 + 2
4
Tricky to see it all written like this, but it works.
2007-07-03 18:28:58 · answer #5 · answered by durhamdouglas 2 · 0⤊ 2⤋
( 3^x +3^-x)^2 - (3^x - 3^-x)^2
=3^2x +3^-2x + 2 - (3^2x + 3^-2x -2)
=4
2007-07-03 18:34:57 · answer #6 · answered by Snoopy 3 · 0⤊ 1⤋
just take the antiderivative of the asymptote to determine the definite integral and use simple logarithm to find the hyperbolic trigonometric functions.
easy as pie
its a Fundamental Theorem of Calculus
2007-07-03 18:14:54 · answer #7 · answered by dr. jay may 1 · 0⤊ 3⤋
(3^2x + 2) - (3^2x - 2)
2 + 2
4
2007-07-03 18:09:01 · answer #8 · answered by cscokid77 3 · 0⤊ 4⤋
Yea it's a very simple math equation. The answer has to do with the simplicities of ditigitisfigureitoutis.
(duh!!!!)-Everyone knows how do answer that
2007-07-03 18:05:51 · answer #9 · answered by Anonymous · 1⤊ 3⤋
YOU NEED TO DO YOUR OWN HOMEWORK TO SUCEED
2007-07-03 18:08:06 · answer #10 · answered by DEVO 1 · 1⤊ 4⤋