math
posted by angiie .
432x63^2
can you tell me how to solve? please

Is 2x6 supposed to be 2 x^6?
What gets squared at the end? The 3 exponent? 
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. 
no its just how i put it. yes

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. 
You need to answer drwls questions above so he/she can help you.

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 
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? 
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. 
If there is no x, then you don't have a variable!
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