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louis413

  • 3 years ago

2(x-1)^3=x+1

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  1. lgbasallote
    • 3 years ago
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    hmm okay?

  2. louis413
    • 3 years ago
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    can u solve it?

  3. louis413
    • 3 years ago
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    ???

  4. lgbasallote
    • 3 years ago
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    solve for x?

  5. louis413
    • 3 years ago
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    yeah

  6. lgbasallote
    • 3 years ago
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    here's my thought let x - 1 = a therefore a +1 = x 2a^3 = a + 1 + 1 2a^3 = a + 2 2a^3 - a - 2 = 0 does that give you any ideas?

  7. louis413
    • 3 years ago
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    2(x-1)^3=x+1 noo sorry not really can u tell me how to do it just by factoring etc?

  8. lgbasallote
    • 3 years ago
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    lol that's what im thinking too

  9. lgbasallote
    • 3 years ago
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    @myininaya pls help here

  10. myininaya
    • 3 years ago
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    Hey @lgbasallote what was wrong with your way?

  11. myininaya
    • 3 years ago
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    It seems to be awesome! :)

  12. lgbasallote
    • 3 years ago
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    i don't know what to do next lol

  13. myininaya
    • 3 years ago
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    Try doing a little of testing possible rational zeros for what you have

  14. myininaya
    • 3 years ago
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    Synthetic division will come in handy

  15. lgbasallote
    • 3 years ago
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    you heard her @louis413 ;)

  16. louis413
    • 3 years ago
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    lol im only in alg/trig in high schooll.. loll wut is that?

  17. myininaya
    • 3 years ago
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    oh wait i think i made a mistake anyways... let me try again but yeah that is what i would do test to see if any possible rational zeros work

  18. myininaya
    • 3 years ago
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    algebra/trig? which one?

  19. louis413
    • 3 years ago
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    well im takling the algebra2/trig course right now at summer school

  20. myininaya
    • 3 years ago
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    synthetic division and looking at possible rational zeros are in algebra

  21. myininaya
    • 3 years ago
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    ok so what have you learned then?

  22. louis413
    • 3 years ago
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    factoring... quadratic formula completeing the squares etc

  23. myininaya
    • 3 years ago
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    And is your equation written correctly?

  24. louis413
    • 3 years ago
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    yes can it be solved like 2(x-1)^2(x-1)-1(X-1)=0

  25. myininaya
    • 3 years ago
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    I can only see to find a real approximation by looking at the graph and seeing where it crosses the x-axis

  26. shubhamsrg
    • 3 years ago
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    would you like to go for a trignometric solution? might seem complicated but is easy.. involves inverse function just a bit..and sin 3x etc formullas ?

  27. shubhamsrg
    • 3 years ago
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    well.. in 2a^3 - a - 2 = 0 let a = m sin y => 2m^3 sin^3 y - msin y -2 =0 multiplying both sides by (-2/m^3) => -4sin^3 y + 2sin y/(m^2) + 4/(m^3) =0 our aim is to convert this in the form of -4sin^3 y + 3 siny which will convert it into sin 3y and simplify things we have to chose a particular value of m for that purpose in our case 2/m^2 = 3 or m = sqrt(3/2) and -sqrt(3/2) thus we have these 2 eqns : -4sin^3 y + 3sin y + 8sqrt(2)/(3sqrt(3)) = 0 and -4sin^3 y + 3sin y - 8sqrt(2)/(3sqrt(3)) = 0 => sin 3y = - 8sqrt(2)/(3sqrt(3)) and =>sin 3y = 8sqrt(2)/(3sqrt(3)) now using any scientific calculator, we may find value of y,then a(=msin y) and then finally x(=a+1)

  28. shubhamsrg
    • 3 years ago
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    hope i was able to explain well..

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