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Definite integrals and finding "z"



I identified that x^2 would become

But even with me substituing x for z; to give z^3/3-9=0

I remain unsure how to proceed further.
(edited 9 years ago)
Original post by apronedsamurai


I identified that x^2 would become

But even with me substituing x for z; to give z^3/3-9=0

I remain unsure how to proceed further.


Now solve for Z.
Original post by apronedsamurai


I identified that x^2 would become

But even with me substituing x for z; to give z^3/3-9=0

I remain unsure how to proceed further.


Is there an actual question
Reply 3
Avatar for User947387
User947387
OP
17
Original post by uberteknik
Now solve for Z.



I thought it was z^3/3=9

I then just took an educated guess; that it was 3; because 3^3=27/3=9

But that was more guesswork and intuition, I am wondering is this like approximiate roots, i.e. you use the trial and error method, or is there a more concrete method/trick to be used
Original post by apronedsamurai
I thought it was z^3/3=9

I then just took an educated guess; that it was 3; because 3^3=27/3=9

But that was more guesswork and intuition, I am wondering is this like approximiate roots, i.e. you use the trial and error method, or is there a more concrete method/trick to be used


Why would you need to guess

Z^3/3 = 9

Multiplying both sides by 3 gives

Z^3 = 27

Cube rooting both sides gives

Z = 3
Reply 5
Avatar for User947387
User947387
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17
I wasn't sure that was the approach to take, that is why I am asking....
Reply 6
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Phichi
11
Original post by apronedsamurai
I wasn't sure that was the approach to take, that is why I am asking....


TenofThem gave you the approach, it's simple manipulation, multiplying, taking cube roots etc etc. You're trying to isolate the z term.
Reply 7
Avatar for davros
davros
16
Original post by apronedsamurai
I thought it was z^3/3=9

I then just took an educated guess; that it was 3; because 3^3=27/3=9

But that was more guesswork and intuition, I am wondering is this like approximiate roots, i.e. you use the trial and error method, or is there a more concrete method/trick to be used


There's no "guesswork" involved in this - it's just basic GCSE rearrangement.

If z33=9\dfrac{z^3}{3} = 9 then z3=27z^3 = 27 and so z=273=3z = \sqrt[3]{27} = 3

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