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Find the values of a for real and distinct roots

  • #1
utkarshakash
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Homework Statement


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

Homework Equations



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?
 

Answers and Replies

  • #2
tiny-tim
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hi utkarshakash! :smile:
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
if you expand it, isn't it just ax2 = a polynomial in x2? :wink:
 
  • #3
utkarshakash
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hi utkarshakash! :smile:


if you expand it, isn't it just ax2 = a polynomial in x2? :wink:
But the fourth degree term is still there.
 
  • #4
tiny-tim
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yes, ax2 = a quadratic polynomial in x2 :smile:
 
  • #5
ehild
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But the fourth degree term is still there.
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:
  • #6
Dick
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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
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.
 
  • #7
ehild
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Unbelievable, Dick! :smile: x^4+x^2+1=(x^2+x+1)(x^2-x+1)

ehild
 
  • #8
Dick
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Unbelievable, Dick! :smile: x^4+x^2+1=(x^2+x+1)(x^2-x+1)

ehild
Oh, its not exactly unbelievable. I used maxima to put everything together and then factored it and noticed that. I cheated.
 
  • #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
 
  • #10
utkarshakash
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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.
Thanks. Finally got the answer. Your method was best.
 
  • #11
utkarshakash
Gold Member
855
13
Can you guys please help me out on my other questions?
 

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