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aceace

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?

  • 9 months ago
  • 9 months ago

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  1. xlegendx
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    when c = 0? hmm

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

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

    • 9 months ago
  4. aceace
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    how did you get that?

    • 9 months ago
  5. xlegendx
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    get what?

    • 9 months ago
  6. aceace
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    your answers

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

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

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

    • 9 months ago
  10. experimentX
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    no I am not who said?

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

    • 9 months ago
  12. aceace
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    well a 99 smart score speaks for itself ;)

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

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

    • 9 months ago
  15. aceace
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    is it just c^2=a^2

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

    • 9 months ago
  17. aceace
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    so is c^2=a^2 a condition?

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

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

    • 9 months ago
  20. experimentX
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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.

    • 9 months ago
  21. hartnn
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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

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

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

    • 9 months ago
  24. experimentX
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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.

    • 9 months ago
  25. experimentX
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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.

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

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

    • 9 months ago
  28. experimentX
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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

    • 9 months ago
  29. experimentX
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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'

    • 9 months ago
  30. experimentX
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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.

    • 9 months ago
  31. aceace
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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

    • 9 months ago
  32. experimentX
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    damn!! this looks like more hard way now.

    • 9 months ago
  33. hartnn
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    mine is equally harder :\

    • 9 months ago
  34. experimentX
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    let me try this on copy

    • 9 months ago
  35. sauravshakya
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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)

    • 9 months ago
  36. aceace
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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?

    • 9 months ago
  37. helder_edwin
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    simple distribution and factorization

    • 9 months ago
  38. sauravshakya
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    yes

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

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

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

    • 9 months ago
  42. sauravshakya
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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

    • 9 months ago
  43. sauravshakya
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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

    • 9 months ago
  44. helder_edwin
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    i agree with the first one

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

    • 9 months ago
  46. sauravshakya
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    oh cant use that when there is inequality

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

    • 9 months ago
  48. helder_edwin
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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

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

    • 9 months ago
  50. ganeshie8
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    how we come up with det >= 0 ?

    • 9 months ago
  51. sauravshakya
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    Because I assumed y is real

    • 9 months ago
  52. ganeshie8
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    ohk we good to assume that ?

    • 9 months ago
  53. helder_edwin
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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

    • 9 months ago
  54. sauravshakya
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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

    • 9 months ago
  55. helder_edwin
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    i was gonna say that.

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

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

    • 9 months ago
  58. ganeshie8
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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

    • 9 months ago
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