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shubhamsrg

  • 2 years ago

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..

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  1. shubhamsrg
    • 2 years ago
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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

  2. shubhamsrg
    • 2 years ago
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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..

  3. shubhamsrg
    • 2 years ago
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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)

  4. hba
    • 2 years ago
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    What is this method called ?

  5. shubhamsrg
    • 2 years ago
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    i dont know/remember.. :|

  6. hba
    • 2 years ago
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    k

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

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

  9. hba
    • 2 years ago
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    let me start then :)

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

  11. shubhamsrg
    • 2 years ago
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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

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

  13. hba
    • 2 years ago
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    What next ?

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

  15. hba
    • 2 years ago
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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 ?

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

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

  18. shubhamsrg
    • 2 years ago
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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?

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

  20. hba
    • 2 years ago
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    Do i expand them using formula ?

  21. shubhamsrg
    • 2 years ago
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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

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

  23. shubhamsrg
    • 2 years ago
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    so,,you're comfortable with the rest ?

  24. hba
    • 2 years ago
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    Yeah i was confused in this part lol :D

  25. shubhamsrg
    • 2 years ago
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    kk.glad i could help..

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

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

  28. shubhamsrg
    • 2 years ago
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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 ?

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

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

  31. shubhamsrg
    • 2 years ago
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    ZZZZZzzz...

  32. hba
    • 2 years ago
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    @shubhamsrg Please.

  33. shubhamsrg
    • 2 years ago
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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..

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

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

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

  37. hba
    • 2 years ago
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    If you feel lazy do it after sometime :)

  38. shubhamsrg
    • 2 years ago
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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..

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

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

  41. shubhamsrg
    • 2 years ago
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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..

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

  43. shubhamsrg
    • 2 years ago
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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|

  44. shubhamsrg
    • 2 years ago
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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

  45. shubhamsrg
    • 2 years ago
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    hope that aids to the lengthiness ..

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

  47. inkyvoyd
    • 2 years ago
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    cardano is better. except casus irriduciblis. Then this is.

  48. shubhamsrg
    • 2 years ago
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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..

  49. Mathmuse
    • 2 years ago
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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.

  50. shubhamsrg
    • 2 years ago
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    that was interesting @mathmuse ..thanks for sharing

  51. mukushla
    • 2 years ago
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    nice one :)

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

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