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aceace

  • one year ago

VERY INTERESTING QUESTION!!!! Find the condition that the equations ax^3+bx+c=0, and cx^3 +bx^2 +a=0 may have a common root?

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  1. xlegendx
    • one year ago
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    when c = 0? hmm

  2. aceace
    • one year ago
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    I would appreciate if someone could solve this cause its been bugging me for days

  3. xlegendx
    • one year ago
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    or maybe if b = 0 and a & c are perfect cubes

  4. aceace
    • one year ago
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    how did you get that?

  5. xlegendx
    • one year ago
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    get what?

  6. aceace
    • one year ago
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    your answers

  7. xlegendx
    • one year ago
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    you were asking for conditions so i was just doing trial and error

  8. aceace
    • one year ago
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    Could someone plz do this with full working thanks :D

  9. aceace
    • one year ago
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    @experimentX you seem like a genius, plz help

  10. experimentX
    • one year ago
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    no I am not who said?

  11. aceace
    • one year ago
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    I have asked heaps of people but no one seems to know how to do it

  12. aceace
    • one year ago
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    well a 99 smart score speaks for itself ;)

  13. experimentX
    • one year ago
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    just compare the coefficient (ax^3+bx+c)/a=0, and (cx^3 +bx^2 +a)/c=0

  14. aceace
    • one year ago
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    ok what do i do when comparing the coefficients for the monic polynomial?

  15. aceace
    • one year ago
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    is it just c^2=a^2

  16. experimentX
    • one year ago
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    the condition that they have same roots is that they must be same function or multiple of same function.

  17. aceace
    • one year ago
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    so is c^2=a^2 a condition?

  18. experimentX
    • one year ago
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    woops!! looks like i misunderstood question ... how many common roots?

  19. aceace
    • one year ago
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    just says a common root, so 1 i guess?

  20. experimentX
    • one year ago
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    let the common root be p ... q,r,s,t be other roots f(x) = a(x-p)(x-q)(x-r) g(x) = c(x-p)(x-s)(x-t) equating to zero, you get ax^3 - a(p+q+r)x^2 + (pq+qr+rp)ax + pqra = ax^3+bx+c do the same for other, eliminate q, and r comparing coefficients and from other eliminate s and t, from both equations you get interms of p, equate 'p' from both and you have your condition. have to see if this get's complicated or not.

  21. hartnn
    • one year ago
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    let that common root be 'y' ay^3+by+c=0, and cy^3 +by^2 +a=0 eliminating 'y' from these two ! O.o

  22. experimentX
    • one year ago
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    yeah that should work too ... solving cubic is real trouble though

  23. aceace
    • one year ago
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    what do you mean by eliminating e.g. like hartnn said eliminating 'y' from these two

  24. experimentX
    • one year ago
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    the condition depends on your initial parameters like 'a', 'b', 'c'. so you have to reduce it to the form in a,b,c which it must satisfy in order to have a common root. you may use http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method to solve for cubic. just find one root and keep squaring or cubic unitl you get rid of surds.

  25. experimentX
    • one year ago
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    if hartnn method didn't work then you can try mine, atmost you have to solve (two) quadratic equation ... i think.

  26. aceace
    • one year ago
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    how do i reduce it to the form in a,b,c?

  27. aceace
    • one year ago
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    soz, i get what you are talking about but i am still confused about how i should start :(

  28. experimentX
    • one year ago
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    that's how i old you ax^3 - a(p+q+r)x^2 + (pq+qr+rp)ax + pqra = ax^3+bx+c p+q+r = 0 pq +qr +rp = b/a pqra = c ------------------ from there equations eliminate q, and r ... you will get an equation in terms of p,a,b,c

  29. experimentX
    • one year ago
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    similarly from here c(x-p)(x-s)(x-t) = cx^3 +bx^2 +a eliminate s and t, you should get equation in p,a,b,d isolate p from both sides .. .this one and another one ... and equal them. you should get the required condition on 'a','b','c'

  30. experimentX
    • one year ago
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    woops!! there has been slight error ax^3 - a(p+q+r)x^2 + (pq+qr+rp)ax - pqra = ax^3+bx+c there might be more erros, just take care of it.

  31. aceace
    • one year ago
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    how does this affect the three equations? and also when i eliminated the q and r i got a cubic of p

  32. experimentX
    • one year ago
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    damn!! this looks like more hard way now.

  33. hartnn
    • one year ago
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    mine is equally harder :\

  34. experimentX
    • one year ago
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    let me try this on copy

  35. sauravshakya
    • one year ago
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    let the common root be y then, ay^3+by+c=0 ....i cy^3+by^2+a=0 ...ii subtract ii from i, (a-c)y^3 + b(y-y^2) + (c-a)=0 (a-c)(y^3 -1) + by(1-y)=0 (a-c)(y-1)(y^2+y+1)-by(y-1)=0 (a-c) (y^2+y+1) -by=0 (a-c)y^2 + (a-c-b)y +(a-c)=0 Now, y exists only if (a-c-b)^2 >= 4*(a-c)*(a-c)

  36. aceace
    • one year ago
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    how did you get from (a-c) (y^2+y+1) -by=0 to (a-c)y^2 + (a-c-b)y +(a-c)=0?

  37. helder_edwin
    • one year ago
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    simple distribution and factorization

  38. sauravshakya
    • one year ago
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    yes

  39. helder_edwin
    • one year ago
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    but why did u omit (y-1)? u cannot just divide by it

  40. sauravshakya
    • one year ago
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    to simplify the expression (y-1) was omitted

  41. helder_edwin
    • one year ago
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    i didn't say it doesn't. but u r excluding y=1 as a possible root.

  42. sauravshakya
    • one year ago
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    oh I missed that...... when y=1 from equation (i) we get a+b+c=0 So, a+b+c=0 also works

  43. sauravshakya
    • one year ago
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    Either a+b+c=0 OR (a-b-c)^2>=4(a-c)^2 (a-b-c)>=2(a-c) a+b-c>=0

  44. helder_edwin
    • one year ago
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    i agree with the first one

  45. helder_edwin
    • one year ago
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    but \[ (-3)^2\geq(-2)^2\Rightarrow -3\geq-2 \] ?

  46. sauravshakya
    • one year ago
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    oh cant use that when there is inequality

  47. sauravshakya
    • one year ago
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    So, either a+b+c=0 OR (a-b-c)^2 >= 4(a-c)^2

  48. helder_edwin
    • one year ago
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    yes. and the best i could do with your inequality is this: \[ b^2-2b(a-c)\geq3(a-c)^2 \] which doesn't look any better than yours

  49. ganeshie8
    • one year ago
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    im trying to understand the solution... in the q they dint say common root needs to be real right ?

  50. ganeshie8
    • one year ago
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    how we come up with det >= 0 ?

  51. sauravshakya
    • one year ago
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    Because I assumed y is real

  52. ganeshie8
    • one year ago
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    ohk we good to assume that ?

  53. helder_edwin
    • one year ago
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    cubic equations have at least on real root. if we assume that it is their real roots that r equal then we r ok

  54. sauravshakya
    • one year ago
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    They cannot only have only one common root imaginery...... if one common root is imaginery then both imaginery roots must be common. But not sure how this would work

  55. helder_edwin
    • one year ago
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    i was gonna say that.

  56. ganeshie8
    • one year ago
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    yea 3 cases :- 1 common root (real) 2 common roots (imaginary) 3 common roots (1 real 2 imaginary)

  57. helder_edwin
    • one year ago
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    in this case y=1 is no longer a problem. so we would go straight into the inequality \[ (a-b-c)^2<4(a-c)^2 \]

  58. ganeshie8
    • one year ago
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    that condition is oly a necessary condition, its not a sufficient condition for ex, (a, b, c) = (1, 3, 2) wont give any common roots

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