# Find the values of a for real and distinct roots

1. Nov 1, 2012

### utkarshakash

1. The problem statement, all variables and given/known data
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

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

2. Nov 1, 2012

### tiny-tim

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

3. Nov 1, 2012

### utkarshakash

But the fourth degree term is still there.

4. Nov 2, 2012

### tiny-tim

yes, ax2 = a quadratic polynomial in x2

5. Nov 2, 2012

### ehild

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
6. Nov 2, 2012

### Dick

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. Nov 2, 2012

### ehild

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

ehild

8. Nov 3, 2012

### Dick

Oh, its not exactly unbelievable. I used maxima to put everything together and then factored it and noticed that. I cheated.

9. Nov 3, 2012

### 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

10. Nov 3, 2012