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aceace

  • 2 years 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
    • 2 years ago
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    when c = 0? hmm

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

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

  4. aceace
    • 2 years ago
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    how did you get that?

  5. xlegendx
    • 2 years ago
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    get what?

  6. aceace
    • 2 years ago
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    your answers

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

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

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

  10. experimentX
    • 2 years ago
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    no I am not who said?

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

  12. aceace
    • 2 years ago
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    well a 99 smart score speaks for itself ;)

  13. experimentX
    • 2 years 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
    • 2 years ago
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    ok what do i do when comparing the coefficients for the monic polynomial?

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

  16. experimentX
    • 2 years 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
    • 2 years ago
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    so is c^2=a^2 a condition?

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

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

  20. experimentX
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    yeah that should work too ... solving cubic is real trouble though

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

  24. experimentX
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    how do i reduce it to the form in a,b,c?

  27. aceace
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    damn!! this looks like more hard way now.

  33. hartnn
    • 2 years ago
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    mine is equally harder :\

  34. experimentX
    • 2 years ago
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    let me try this on copy

  35. sauravshakya
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    simple distribution and factorization

  38. sauravshakya
    • 2 years ago
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    yes

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

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

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

  42. sauravshakya
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    i agree with the first one

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

  46. sauravshakya
    • 2 years ago
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    oh cant use that when there is inequality

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

  48. helder_edwin
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    how we come up with det >= 0 ?

  51. sauravshakya
    • 2 years ago
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    Because I assumed y is real

  52. ganeshie8
    • 2 years ago
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    ohk we good to assume that ?

  53. helder_edwin
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    i was gonna say that.

  56. ganeshie8
    • 2 years 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
    • 2 years 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
    • 2 years 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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