$x^4-2x^3+mx^2-2x+1=0$, $m \in R$, $x_1=x_2 \in R-\{-1\}$

Question

Given the equation: $x^4-2x^3+mx^2-2x+1=0$, $m \in R$

To which interval does $m$ belong, so that the equation has $x_1=x_2 \in \mathbb R\setminus\{-1\}$, provided that $x_1,x_2,x_3,x_4$ are the roots of the equation.

Since I got this from Vieta's:

$$2x_1+x_3+x_4=2$$ $$x_1^2+2x_1x_3+2x_1x_4+x_3x_4=m$$ $$x_1^2x_3x_4=1$$ $$x_1^2x_3+x_1^2x_4+x_1x_3x_4+x_1x_3x_4=2$$

I figured I could do this:

$$x_1^2(x_3+x_4)+2x_1x_3x_4=2$$ $$x_3+x_4=2-2x_1$$ $$x_1x_3x_4= \frac{1}{x_1}$$ $$x_1^2(2-2x_1)+\frac{2}{x_1}=2$$ $$2x_1^4-2x_1^3+2x_1-2=0$$

So I use Horner's on the last line above and get $x_ 1=1$ the only real root which is also $x_2$.

Thus I am able to find $m=2$ which belongs to the interval of the correct answer[which is $(1, \infty )$].

The thing is I am not sure this is the only $m$ that provides a solution to the exercise... Is it the only value of $m$?

Answer 1

hint: since $$x=0$$ is not a solution then write your equation in the form $$x^2+\frac{1}{x^2}-2\left(x+\frac{1}{x}\right)+m=0$$ now set $$x+\frac{1}{x}=t$$ and you will get a quadratic equation in $t$

Answer 2

We can get your result also by the following way. $$(x^4-2x^3+mx^2-2x+1)'=0$$ or $$4x^3-6x^2+2mx-2=0$$ or $$2x^4-3x^3+mx^2-x=0$$ or $$mx^2=-2x^4+3x^3+x,$$ which gives $$x^4-2x^3+(-2x^4+3x^3+x)-2x+1=0$$ or $$x^4-x^3+x-1=0$$ or $$(x^3+1)(x-1)=0,$$ which gives $x=1$ and $m=2$.