Asked by angiie

4|3-2x6|-3^2
can you tell me how to solve? please

Answers

Answered by drwls
Is 2x6 supposed to be 2 x^6?
What gets squared at the end? The -3 exponent?
Answered by helper
Is this your problem?

4|3 - 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.

Re-check your problem.
Answered by angiie
no its just how i put it. yes
Answered by MathMate
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.
Answered by helper
You need to answer drwls questions above so he/she can help you.

Answered by angiie
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
Answered by angiie
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?

Answered by MathMate
Assuming you mean:
4|3-2x^6|-3²=0

Then
4|3-2x^6|=3²
|3-2x^6|=9/4
There are two cases,
a. when 3-2x^6 >0
|3-2x^6|=9/4
=> 3-2x^6=9/4
=> 2x^6=3-9/4=3/4
=> x^6=3/8
=> x=±(3/8)^(1/6) (x∈ℝ, i.e. x is real)

b. when 3-2x^6<0
|3-2x^6|=9/4
=> -(3-2x^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.
Answered by MathMate
If there is no x, then you don't have a variable!
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