What is the best way to solve an equation of the form $(f(x))^2-a(f(x))+b=x$?

Question

On a math contest I was told to solve the equation $$(x^2-3x+1)^2-3(x^2-3x+1)+1=x$$

For this particular problem I simplified by letting $$a\equiv x^2-3x+1$$

Then I continued with $$a^2-3a+1-x=0$$ $$a=\frac{3\pm\sqrt{9-4\left(1-x\right)}}{2}$$ $$3\pm\sqrt{5+4x}=2a=2x^2-6x+2$$ $$\pm\sqrt{5+4x}=2x^2-6x-1$$ I am not sure how to finish off this problem. Also, I have seen a bunch of problems like this in the past. What is the best way to approach a problem like this and also how could I finish solving this problem? Could someone also explain why f(f(x))=x has the same solutions as f(x)=x.

Answer 1

Add $x^2-3x+1$ to both sides, giving $$(x^2-3x+1)^2 - 2(x^2-3x+1) + 1 = x + x^2-3x+1 = x^2-2x+1,$$ or $$((x^2-3x+1)-1)^2 = (x-1)^2.$$ Thus $(x^2-3x+1)-1 = \pm (x-1)$. The plus sign gives $$x^2-4x+1=0\quad\Rightarrow\quad x = 2\pm\sqrt{3}.$$ The minus sign gives $$x^2-2x-1=0\quad\Rightarrow\quad x = 1\pm\sqrt{2}.$$

Answer 2

The question is about $f(x)=x^2-3x+1$, and solutions to $f(f(x))=x$.
Two solutions to $f(f(x))=x$ are the solutions to $f(x)=x$.
I'm not sure why you get a cubic instead of a quartic polynomial.

Answer 3

Let $f(x) = x^2-3x+1$.

Idea

The important observation here is following: If you have a solution $a$ that satisfies $f(a) = a$ then obviously $a$ also solves the original equation because $f(f(a)) = f(a) = a$.

Finding the $a$ that solve $f(a) = a$ is easy, that is just a quadratic equation. This means you can divide $f(f(x))-x$ by $(x-a)$ for both solutions $a$ and you get a quadratic equation in $x$ which again can be solved easily.

Calculations

So lets first solve $f(x) = x$ This is aequivalent to $0 = x^2-4x+1$ which has the solutions $x_{1,2} = 2 \pm \sqrt{3}$

Now we can divide $f(f(x))-x$ by $(x-x_1)$ and $(x-x_2)$ or combined by $(x-x_1)(x-x_2) = x^2-4x+1 $.

Via polynomial division we get $$(f(f(x))-1)/(x^2-4x+1) = x^2-2x+1$$

Remember taht $f(f(x))-x = 0$ is equivalent now to

$0=f(f(x))-x=(x-x_1)(x-x_2)(x^2-2x+1) = f(f(x))-x$.

As $x^2-2x+1$ has the zeros $x_{3,4} = 1 \pm \sqrt{2}$ we can further factorize

$0=f(f(x))-x=(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ and we have all four solutions.

Answer 4

HINT: simplifying and factorizing the given equation we have to solve $$\left( {x}^{2}-4\,x+1 \right) \left( {x}^{2}-2\,x-1 \right) =0$$ it is easy now

Answer 5

As already noted, this is a quartic equation of the form $f(f(x)) - x = 0$ where $f$ is a polynomial. Then the factor $f(x) - x$ must divide the expression $f(f(x)) - x$. This will allow you to factor the quartic term into two quadratic terms both of which can be factored using the usual formula. In this particular case, $f(x) - x = x^2-4x+1$ and you can write $$ f(f(x)) - x = (x^2-3x+1)^2 - 3 (x^2-3x+1) + 1 - x = (x^2-4x+1)(x^2-2x-1) $$