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August 5, 2015

August 5, 2015

Posted by **angiie** on Monday, February 7, 2011 at 10:32pm.

can you tell me how to solve? please

- math -
**drwls**, Monday, February 7, 2011 at 10:36pmIs 2x6 supposed to be 2 x^6?

What gets squared at the end? The -3 exponent?

- math -
**helper**, Monday, February 7, 2011 at 10:39pmIs 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.

- math -
**angiie**, Monday, February 7, 2011 at 10:40pmno its just how i put it. yes

- multiple posts -
**MathMate**, Monday, February 7, 2011 at 10:42pmangiie,

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 10:46pmYou need to answer drwls questions above so he/she can help you.

- math -
**angiie**, Monday, February 7, 2011 at 10:47pmMathMate

sorry i didnt 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 10:51pmto 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 10:55pmAssuming 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.

- math -
**MathMate**, Monday, February 7, 2011 at 10:56pmIf there is no x, then you don't have a variable!