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shubhamsrg

TUTORIAL : on how to solve a 3 degree eqn in x using trigonometry .. i learnt this method quite some time ago and maybe everyone should give this a reading..

  • one year ago
  • one year ago

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  1. shubhamsrg
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    we have : ax^3 + bx^2 + cx+ d =0 for simplification, i like to keep coefficient of x^3 = 1 for that,,lets divide both sides by a (given a not equal to 0) lets say b/a =p , c/a = q , d/a =r => x^3 + px^2 + qx +r =0 what we aim here firstly is to convert this into depressed form ( make co-efficient of 2nd highest degree =0) lets substitute x = (y + m) , for some new variable y and a constant m , we will understand the purpose of doing this soon.. => (y+m)^3 + p(y+m)^2 + q(y+m) + r =0 =>y^3 + y^2(p + 3m) + y(2m + 3m^2 +q) + (m^3 + pm^2 + qm +r) =0 as i said earlier,we want p+ 3m =0 => m= -p/3 thus on making that substitution, we'll get y^3 + Ay + B = 0 where A and B are some known constants.. now if we calculate y , we can easily calculate x as x= y+m or x= y -p/3 here comes the trignometric part: this glorious formulla will do wonders for us as you'll find below : sin 3@ = 3sin @ - 4sin^3 @ (@ denotes theta) so we have y^3 + Ay + B=0 let y = t sin@ for some constant t , => t^3 sin^3 @ + At sin@ + B =0 lets multiply both sides by -4/t^3 => -4 sin^3 @ + -4A sin@ /t^2 + -4B/t^3 = 0 we have to chose t such that coefficient of sin@ = 3 => -4A/t^2 = 3 =>t = sqrt( -4A/3) and -sqrt( -4A/3) on making that substitution, we see we have -4sin^3 @ + 3sin@ =C for some new constant C. => sin3@ = C =>3@ = sin^-1 (C) @ = 1/3 * sin^-1 (C) since you know the value of @, you then know value of y = t sin@ and then x = y+m

    • one year ago
  2. shubhamsrg
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    all this might seem tedious and long at first,,but this is very advantageous to know.. this one yields both complex roots as well as real roots..2 of them .. (since you get 2 values of t) hope that helps..

    • one year ago
  3. shubhamsrg
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    another thing which we can make out from here that, in any eqn like x^n + px^(n-1) + qx^(n-2) ......... + constant term =0 to make it depressed eqn , we can always substitute x = y - (p/n)

    • one year ago
  4. hba
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    What is this method called ?

    • one year ago
  5. shubhamsrg
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    i dont know/remember.. :|

    • one year ago
  6. hba
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    k

    • one year ago
  7. hba
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    3x^3+2x^2+x+1=0 Show me how to solve this step by step please.

    • one year ago
  8. shubhamsrg
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    you do it and i shall add my words in between when required..

    • one year ago
  9. hba
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    let me start then :)

    • one year ago
  10. hba
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    Do i have to suppose x^3=1 ?

    • one year ago
  11. shubhamsrg
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    nops..i meant coefficient of x^3 should be made 1,,just for simplification.. so you may divide both sides by 3

    • one year ago
  12. hba
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    Dividing the whole equation by 3 x^3+2/3 x^2+1/3x+1/3=0

    • one year ago
  13. hba
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    What next ?

    • one year ago
  14. shubhamsrg
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    x = y - p/3 .. make that substitution,, p = 2/3 here..

    • one year ago
  15. hba
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    p=2/3,q=1/3,r=1/3 I did not get the substitution part lol ? Do you think this method is more effective then the others ?

    • one year ago
  16. shubhamsrg
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    i cant say that,,since i dont know other methods! :P nevermind,,whats the problem is substituting ?

    • one year ago
  17. shubhamsrg
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    like in x+2 = y, you can substitute directly x=2 ,, similarly,,substitute x= y - p/3

    • one year ago
  18. shubhamsrg
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    i'll give a kick.. x^3+2/3 x^2+1/3x+1/3=0 (y - 2/9)^3+ 2/3 (y- 2/9)^3 + 1/3 ( y- 2/9) + 1/3 = 0 did you follow?

    • one year ago
  19. hba
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    Ok so we have just done a simple substitution here x=y-2/9.

    • one year ago
  20. hba
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    Do i expand them using formula ?

    • one year ago
  21. shubhamsrg
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    yep..expand and simplify .. write what you get .. sorry am too lazy to check your work or do it myself..so be careful..please dont mind. :P

    • one year ago
  22. hba
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    Anyways i understood the part where i was confused,Thanks a lot.

    • one year ago
  23. shubhamsrg
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    so,,you're comfortable with the rest ?

    • one year ago
  24. hba
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    Yeah i was confused in this part lol :D

    • one year ago
  25. shubhamsrg
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    kk.glad i could help..

    • one year ago
  26. Hero
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    Is there a textbook or video tutorial I can find this in?

    • one year ago
  27. hba
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    If i remember the name of the method i could have told you :(

    • one year ago
  28. shubhamsrg
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    well one of my seniors had told me this..i really cant tell you any name or any vids or theories about it..sorry.. anyways,,are you stuck somewhere ?

    • one year ago
  29. hba
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    @Hero This method does not intrest me :) Its very long.

    • one year ago
  30. shubhamsrg
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    sorry buddy,,too lazy doing all that..though i could help you ..but giving examples..nah,,not gonna do that..

    • one year ago
  31. shubhamsrg
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    ZZZZZzzz...

    • one year ago
  32. hba
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    @shubhamsrg Please.

    • one year ago
  33. shubhamsrg
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    i dont regret that..i made an effort..and it was satisfactory for me..i dont mind your foolish reviews since you ofcorse,dont get it..

    • one year ago
  34. hba
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    @shubhamsrg If you give an example maybe you would be getting more medals think about that :D

    • one year ago
  35. shubhamsrg
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    ohh yes,,the medals..it was always my dream to earn more and more medals on open study.. !! o.O

    • one year ago
  36. hba
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    Oh come on just provide us with an example please ?

    • one year ago
  37. hba
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    If you feel lazy do it after sometime :)

    • one year ago
  38. shubhamsrg
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    nevermind,,not gonna do that..if you dont get it..put up your query here..i may help/improvise// but i hate doing hard work more often than not..

    • one year ago
  39. RajshikharGupta
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    it is tooo lengthy but indeed beautiful we use to call it 'MAJDOORI'

    • one year ago
  40. shubhamsrg
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    we?? o.O who all do you represent? :P

    • one year ago
  41. shubhamsrg
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    alright,,seems its lengthy.. in that case,,gimme some time,,i'll post a general formula to the value of x directly..

    • one year ago
  42. sirm3d
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    except for the trigonometric substitution, solving a cubic equation appeared in Ars Magna of Girolamo Cardamo, or Cardan.

    • one year ago
  43. shubhamsrg
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    here's the general formula for solution of x in x^3 + px^2 + qx + r =0 using trig. substitution : |dw:1354008782695:dw|

    • one year ago
  44. shubhamsrg
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    where: m = sqrt(-4A /3) and -sqrt(-4A/3) A = q- (p^2 /3) B = (2p^3 /27) + (qp /3) + r

    • one year ago
  45. shubhamsrg
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    hope that aids to the lengthiness ..

    • one year ago
  46. shubhamsrg
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    sorry, correction : that'll be -p/3 not +p/3 there in the drawing.. rest all the same..

    • one year ago
  47. inkyvoyd
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    cardano is better. except casus irriduciblis. Then this is.

    • one year ago
  48. shubhamsrg
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    ohh..i just googled over!! that was amazing! thanks @inkyvoyd also,,just found out vieta's substitution method..too good..probably the easiest..

    • one year ago
  49. Mathmuse
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    Haha, reading about this and found this excerpt: The long-sought solution of the general cubic was found, in 1535 by Niccol`o Tartaglia (c. 1500-57). This achievement brought him great celebrity, and he spent the next 10 years visiting the crownedheads of Europe and solving their cubics for them. However, he was persuaded to divulge his secret, on the promise of complete confidentiality, by Girolamo Cardano (1501–76), who promptly published it in his algebra book The Great Art. There followed an acrimonious dispute between Tartaglia and Cardano which preoccupied much of Tartaglia’s later life. At one point, the two agreed to a public duel, in which they would each bring along their favourite mathematical problems for the other to solve. However, it never took place @Hero http://www.admissionstests.cambridgeassessment.org.uk/adt/digitalAssets/110501_Advanced_Problems_in_Mathematics.pdf question #12 has an example of this.

    • one year ago
  50. shubhamsrg
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    that was interesting @mathmuse ..thanks for sharing

    • one year ago
  51. mukushla
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    nice one :)

    • one year ago
  52. shubhamsrg
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    aha..thanks..your appreciations means a lot to me sir.. :)

    • one year ago
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