# Find the values of a for real and distinct roots

Gold Member

## 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

## 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?

Related Precalculus Mathematics Homework Help News on Phys.org
tiny-tim
Homework Helper
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? Gold Member
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
Homework Helper
yes, ax2 = a quadratic polynomial in x2 ehild
Homework Helper
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:
Dick
Homework Helper
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
Homework Helper
Unbelievable, Dick! x^4+x^2+1=(x^2+x+1)(x^2-x+1)

ehild

Dick
Homework Helper
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
Homework Helper
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

Gold Member
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.