Posted by angiie on Monday, February 7, 2011 at 10:32pm.
432x63^2
can you tell me how to solve? please

math  drwls, Monday, February 7, 2011 at 10:36pm
Is 2x6 supposed to be 2 x^6?
What gets squared at the end? The 3 exponent? 
math  helper, Monday, February 7, 2011 at 10:39pm
Is this your problem?
43  2x^6  3^2 = ?
I think you must have copied this wrong.
This problem, as typed, is not easy to solve.
I used an online calculator and there are 12 different roots for x (4 real and 10 complex.
Recheck your problem. 
math  angiie, Monday, February 7, 2011 at 10:40pm
no its just how i put it. yes

multiple posts  MathMate, Monday, February 7, 2011 at 10:42pm
angiie,
Please refrain from multiple posts. The time for teachers to read multiple posts or to determine if they are multiple posts could very well be used for answering questions. 
math  helper, Monday, February 7, 2011 at 10:46pm
You need to answer drwls questions above so he/she can help you.

math  angiie, Monday, February 7, 2011 at 10:47pm
MathMate
sorry i didn't know it did that. my commputer was taking a while to load and i clicked a few times before i let it load. sorry agaiin 
math  angiie, Monday, February 7, 2011 at 10:51pm
to drwls
My equation was just how i put it. 2x6 is not supposed to be 2 x^6. its just 2x6.
And the 3 exponent yes it gets squared at the end.
what dose these lines mean   squareroot? 
math  MathMate, Monday, February 7, 2011 at 10:55pm
Assuming you mean:
432x^63²=0
Then
432x^6=3²
32x^6=9/4
There are two cases,
a. when 32x^6 >0
32x^6=9/4
=> 32x^6=9/4
=> 2x^6=39/4=3/4
=> x^6=3/8
=> x=±(3/8)^(1/6) (x∈ℝ, i.e. x is real)
b. when 32x^6<0
32x^6=9/4
=> (32x^6)=9/4
=> 2x^6=9/4+3=21/4
=> x^6=21/8
=> x=±(21/8)^(1/6) (x∈ℝ)
As Helper said, there are 2*4 complex roots which we do not consider in the real domain. 
math  MathMate, Monday, February 7, 2011 at 10:56pm
If there is no x, then you don't have a variable!