Finding Solutions to a Quartic Equation

[Refraction]

Homework Statement

I've been given the following quartic equation in a question:

x = (1 + 2a + a^2)x - (2a + 3a^2 + a^3)x^2 + (2a^2 + 2a^3)x^3 - a^3x^4

and need to show that two of the solutions (that aren't 0 and 1) can be given by:

x= \frac{(2+a) \pm \sqrt{a^2 - 4}}{2a}

2. The attempt at a solution

I think I have to start by dividing it all by x - when I put this new equation into wolfram alpha it gives the right answer, I just have no idea how to get there (or where to start), since I haven't done a whole lot with cubics or quartics before.

 

[Hurkyl]

I think I have to start by dividing it all by x

Can you say why you think so?

 

[Borek]

Why don't you just put expression given as x into the equation?

 

[Refraction]

Can you say why you think so?

Sorry, I left out a part of the equation above. It was actually x = [what I already gave], I've edited that in now. It just seems like that way it would be easier to rearrange into the form needed, I'm not 100% sure about that though.

 

[Borek]

I thought you mean f(x) = 0. Still, what I wrote above holds.

 

[Refraction]

what did you mean by putting the expression given as x into the equation? Putting it into the second one there/the other way around?

 

[Borek]

Sorry if my English failed.

Substitute value of x given in the second equation for all occurrences of x in the first equation. That will yield equation in "a" - and if second equation really gives solution, equation in "a" should be just an identity.

 

[HallsofIvy]

You are given two putative solutions. Just put them into the equation, do the algebra, and see whether or not they satisfy the equation.

 

[Refraction]

Would there be a reasonably simple way to get to the two results given without substituting them into the first equation? I have a feeling that I'm supposed to do it that way, and not just by replacing x with the answers given.

It seems like I could get to that form with some rearranging and factorising but I'm not sure where to begin with that, or if there's any specific techniques that might help.

 

[Petek]

Go back to your original equation and rewrite in the form f(x) = 0. Observe that you can factor out ax, so your equation looks like axg(x) = 0. Now g(x) is a cubic polynomial and you know that 1 is a root. Use long division to compute g(x)/(x-1). The result is a polynomial of degree 2. Now use the quadratic formula to find its roots. You end up with the result in your first post.

 

[Refraction]

Thanks, that worked perfectly!