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Find the values of a for real and distinct roots

  1. Nov 1, 2012 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Find the values of 'a' so that two of the roots of the equation [itex](a-1)(x^2+x+1)^2=(a+1)(x^4+x^2+1)[/itex] are real and distinct

    2. Relevant equations

    3. The attempt at a solution
    I am thinking of converting this equation in quadratic form so that I can find discriminant and make it greater than 0.
    This is how I started
    Let [itex] x^2+x+1=y[/itex]
    [itex](a-1)y^2=(a+1)(x^4+x^2+1)[/itex]
    But I'm stuck here. The fourth degree term still prevails.
    Is there any way this equation can be converted to a quadratic?
     
  2. jcsd
  3. Nov 1, 2012 #2

    tiny-tim

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    hi utkarshakash! :smile:
    if you expand it, isn't it just ax2 = a polynomial in x2? :wink:
     
  4. Nov 1, 2012 #3

    utkarshakash

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    But the fourth degree term is still there.
     
  5. Nov 2, 2012 #4

    tiny-tim

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    yes, ax2 = a quadratic polynomial in x2 :smile:
     
  6. Nov 2, 2012 #5

    ehild

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    Expand, collect the terms with a and without it,and try to factorise again. The equation becomes quite simple and easy to find those "a" values which ensure real and distinct roots.

    ehild
     
    Last edited: Nov 2, 2012
  7. Nov 2, 2012 #6

    Dick

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    Or just notice that x^2+x+1 is a factor of x^4+x^2+1. Either way its pretty easy to write your equation as the product of two quadratics.
     
  8. Nov 2, 2012 #7

    ehild

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    Unbelievable, Dick! :smile: x^4+x^2+1=(x^2+x+1)(x^2-x+1)

    ehild
     
  9. Nov 3, 2012 #8

    Dick

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    Oh, its not exactly unbelievable. I used maxima to put everything together and then factored it and noticed that. I cheated.
     
  10. Nov 3, 2012 #9

    ehild

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    I tried to write x^2+x+1=(x^3-1)/(x-1) and x^4+x^2+1=(x^6-1)/(x^2-1)

    [tex](a-1)\frac{(x^3-1)^2}{(x-1)^2}=\frac{(x^3-1)(x^3+1)}{(x-1)(x+1)}[/tex] and simplifying (x can not be 1) and still did not notice...

    But it becomes
    [tex](a-1)\frac{(x^3-1)}{(x-1)}=(a+1)\frac{(x^3+1)}{(x+1)}[/tex]
    [tex](a-1)(x^2+x+1)=(a+1)(x^2-x+1)[/tex]

    ehild
     
  11. Nov 3, 2012 #10

    utkarshakash

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    Thanks. Finally got the answer. Your method was best.
     
  12. Nov 3, 2012 #11

    utkarshakash

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    Can you guys please help me out on my other questions?
     
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