+0

# Some hard alg2 question

0
141
3

Find the exact result of:

$$(\sqrt{3}-2)^{2017}*(\sqrt{3}+2)^{2018}$$

edit: my bad, bases aren't the same

Guest Dec 22, 2017
edited by Guest  Dec 22, 2017
Sort:

#1
+2

OK! I thought there was more to it than that!.

The exact result is =-(2 + sqrt(3)) - According to "Mathematica 11 Home Edition"

Note: The steps it uses are soooooooo Loooooooooooong, it is rediculous!! There has to be a simpler way of doing it. Maybe somebody else can come up with a short answer.

Guest Dec 22, 2017
edited by Guest  Dec 22, 2017
#2
+6616
+2

$$(\sqrt3-2)^{2017}(\sqrt3+2)^{2018} \\~\\ =\,(\sqrt3-2)^{2017}(\sqrt3+2)^{2017}(\sqrt 3+2) \\~\\ =\,[\,(\sqrt3-2)(\sqrt3+2)\,]^{2017}\,(\sqrt3+2) \\~\\ =\,[\,3-4\,]^{2017}\,(\sqrt3+2) \\~\\ =\,[-1]^{2017}\,(\sqrt3+2) \\~\\ =\,(-1)(\sqrt3+2) \\~\\ =\,-\sqrt3-2$$

hectictar  Dec 22, 2017
#3
+84387
+2

Nice, hectiictar!!!!

[ This one was a little easier after the edit......LOL!!!  ]

CPhill  Dec 22, 2017

### 15 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details