How to solve $x^4-2x^3-x^2+2x+1=0$?

Question

How to solve $x^4-2x^3-x^2+2x+1=0$?

Answer given is: $$\frac{1+\sqrt5}{2}$$

I tried solving it by taking common factors:

$$x^3(x-2)-x(x-2)+1=0 $$ $$x(x-2)(x^2-1)+1=0 $$ $$(x+1)(x)(x-1)(x-2)+1=0$$

But it's not leading me anywhere.

Answer 1

Another way to notice the factorization $$x^4-2x^3-x^2+2x+1=0$$ Since $x=0$ is not the root of the equation, divide by $x^2$ to get $$x^2 -2x-1 + \frac{2}{x} + \frac{1}{x^2} = 0$$ Rewrite it as $$x^2+\frac{1}{x^2} - 2\left(x-\frac{1}{x}\right) - 1 = 0$$ or $$\left(x-\frac1x\right)^2 + 2 - 2\left(x-\frac{1}{x}\right) - 1 = 0$$ Substitute $t = x - 1/x$ $$t^2 + 2 - 2t - 1 = 0\\ t^2 - 2t + 1 = 0 \\ (t-1)^2 = 0$$ Substitute back to get the final result $$\left(x - \frac{1}{x} - 1\right)^2 = 0$$ which says $$(x^2-x-1)^2 = 0$$

Answer 2

Hints:

Let $x-2=p.$

Then you will have a perfect square. It comes from a general fact:

"One more than a product of four consecutive positive integers is a perfect square."

Please take a look at the following link:

Prove that the product of four consecutive positive integers plus one is a perfect square

Answer 3

General solution:

I will solve a specific quartic equation that is a specific case of a general quartic equation.

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Let, $a≠0,~ b≠0$, then we have

$$ax^4+bx^3+cx^2+dx+e=0$$

$$x^2+\frac e{ax^2}+\frac ba x+\frac d{ax}+\frac ca=0$$

$$\begin{align}&x^2+\frac ea \times \frac 1{x^2}+\frac ba \left(x+\frac {d}{bx}\right)+\frac ca=0&\end{align}$$

This quartic equation can be directly converted to the quadratic equation in the case below, avoiding the cubic equation.

$$\begin{align}&x+\frac {d}{bx}=t \\ \implies &t^2=x^2+\frac{d^2}{b^2}\times \frac {1}{x^2}+\frac{2d}{b}\end{align}$$

Then, if $$\frac ea =\frac{d^2}{b^2}$$

We have

$$t^2-\frac{2d}{b}+\frac ba t+\frac ca=0$$

$$t^2+\frac ba t+\left(\frac ca-\frac{2d}{b}\right)=0$$

The last equation is a quadratic equation.

After solving quadratic, you wil get

$$x+\frac {d}{bx}=t$$

$$bx^2-btx+d=0$$

The last equation is also a quadratic equation.

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In your case we have

$$a=1,b=-2,c=-1, d=2, e=1$$

This means

$$\frac ea =\frac{d^2}{b^2}$$

holds.

Answer 4

$x^4-2x^3-x^2+2x+1 = 0$

You should split the middle terms and try to look for a common factor.

$x^4-x^3-x^2-x^3+x^2+x-x^2+x+1 = 0$

$x^2(x^2-x-1)-x(x^2-x-1)-1(x^2-x-1) = 0$

$(x^2-x-1)^2 = 0$

Take the square root:

$x^2-x-1 = ±0$

$x^2-x-1 = 0 \lor x^2-x-1 = -0$

Add $\frac{5}{4}$:

$x^2-x+\frac{1}{4} = \frac{5}{4} \lor x^2-x+\frac{1}{4} = \frac{5}{4}$

We have two identical quadratic equations so all roots will coincide and there will be two repeated roots instead of four distinct roots.

$(x-\frac{1}{2})^2 = \frac{5}{4}$

Take the square root:

$x-\frac{1}{2} = ±\sqrt{\frac{5}{4}}$

$x-\frac{1}{2} = \sqrt{\frac{5}{4}} \lor x-\frac{1}{2} = -\sqrt{\frac{5}{4}}$

Add $\frac{1}{2}$:

$x = \frac{1}{2} + \frac{\sqrt{5}}{2} \lor x = \frac{1}{2} - \frac{\sqrt{5}}{2}$

Answer 5

obten la raiz negativa de la siguiente funcion con 4 cifras decimales.

$2x^4-2x^3+x^2+3x-4=0 \quad\quad$ newton raphson