Find the values of a for real and distinct roots

[utkarshakash]

Homework Statement

Find the values of 'a' so that two of the roots of the equation (a-1)(x^2+x+1)^2=(a+1)(x^4+x^2+1) 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 x^2+x+1=y
(a-1)y^2=(a+1)(x^4+x^2+1)
But I'm stuck here. The fourth degree term still prevails.
Is there any way this equation can be converted to a quadratic?

 

[tiny-tim]

hi utkarshakash!

Find the values of 'a' so that two of the roots of the equation (a-1)(x^2+x+1)^2=(a+1)(x^4+x^2+1) are real and distinct

if you expand it, isn't it just ax² = a polynomial in x²?

 

[utkarshakash]

hi utkarshakash!


if you expand it, isn't it just ax² = a polynomial in x²?

But the fourth degree term is still there.

 

[tiny-tim]

yes, ax² = a quadratic polynomial in x²

 

[ehild]

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

 

[Dick]

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.

 

[ehild]

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

ehild

 

[Dick]

Unbelievable, Dick! 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.

 

[ehild]

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)

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

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

ehild

 

[utkarshakash]

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.

 

[utkarshakash]

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