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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
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..
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)
What is this method called ?
i dont know/remember.. :|
3x^3+2x^2+x+1=0 Show me how to solve this step by step please.
you do it and i shall add my words in between when required..
let me start then :)
Do i have to suppose x^3=1 ?
nops..i meant coefficient of x^3 should be made 1,,just for simplification.. so you may divide both sides by 3
Dividing the whole equation by 3 x^3+2/3 x^2+1/3x+1/3=0
What next ?
x = y - p/3 .. make that substitution,, p = 2/3 here..
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 ?
i cant say that,,since i dont know other methods! :P nevermind,,whats the problem is substituting ?
like in x+2 = y, you can substitute directly x=2 ,, similarly,,substitute x= y - p/3
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?
Ok so we have just done a simple substitution here x=y-2/9.
Do i expand them using formula ?
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
Anyways i understood the part where i was confused,Thanks a lot.
so,,you're comfortable with the rest ?
Yeah i was confused in this part lol :D
kk.glad i could help..
Is there a textbook or video tutorial I can find this in?
If i remember the name of the method i could have told you :(
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 ?
sorry buddy,,too lazy doing all that..though i could help you ..but giving examples..nah,,not gonna do that..
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..
ohh yes,,the medals..it was always my dream to earn more and more medals on open study.. !! o.O
Oh come on just provide us with an example please ?
If you feel lazy do it after sometime :)
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..
it is tooo lengthy but indeed beautiful we use to call it 'MAJDOORI'
we?? o.O who all do you represent? :P
alright,,seems its lengthy.. in that case,,gimme some time,,i'll post a general formula to the value of x directly..
except for the trigonometric substitution, solving a cubic equation appeared in Ars Magna of Girolamo Cardamo, or Cardan.
here's the general formula for solution of x in x^3 + px^2 + qx + r =0 using trig. substitution : |dw:1354008782695:dw|
where: m = sqrt(-4A /3) and -sqrt(-4A/3) A = q- (p^2 /3) B = (2p^3 /27) + (qp /3) + r
hope that aids to the lengthiness ..
sorry, correction : that'll be -p/3 not +p/3 there in the drawing.. rest all the same..
cardano is better. except casus irriduciblis. Then this is.
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.
nice one :)
aha..thanks..your appreciations means a lot to me sir.. :)