# Solve an equation II

#### anemone

##### MHB POTW Director
Staff member
Solve the equation below for real solutions:

$\dfrac{1}{(x^2+x+2)^2}+\dfrac{1}{16}-\dfrac{1}{x^4-x^3+x^2+3x+4}=0$

Last edited:

#### Boromir

##### Banned
Solve the equation below for real solutions:

$\dfrac{1}{(x^2+x+2)^2}+\dfrac{1}{16}-\dfrac{1}{x^4-x^3+x^2+3x+4}=0$
x=limit of y as y tends to infinity

#### mente oscura

##### Well-known member
Solve the equation below for real solutions:

$\dfrac{1}{(x^2+x+2)^2}+\dfrac{1}{16}-\dfrac{1}{x^4-x^3+x^2+3x+4}=0$
Hello.

Brute force.

$$x^4-x^3+x^2+3x+4<(x^2+x+2)^2$$

$$x^4-x^3+x^2+3x+4<x^4+2x^3+5x^2+4x+4$$

$$3x^3+4x+x>0 \rightarrow{}x>0$$

$$x^4-x^3+x^2+3x+4<16$$

$$x^4-x^3+x^2+3x-12<0$$

$$(x^2-3)(x^2-x+4)<0$$

$$x< \sqrt{3}$$

$$\dfrac{1}{(x^2+x+2)^2}+\dfrac{1}{16}-\dfrac{1}{x^4-x^3+x^2+3x+4}=0$$

$$x^8+x^7+4x^6+4x^5+15x^4+23x^3+36x^2+28x+16=16(3x^3+4x^2+x)$$

$$x^8+x^7+4x^6+4x^5+15x^4-25x^3-28x^2+12x+16=0$$

Adding the coefficients, it is "0". Therefore x=1 it's root.

$$\dfrac{x^8+x^7+4x^6+4x^5+15x^4-25x^3-28x^2+12x+16}{x-1}=$$

$$=x^7+2x^6+6x^5+10x^4+25x^3-28x-16$$

Adding the coefficients, it is "0". Therefore x=1 it's root.

$$\dfrac{x^7+2x^6+6x^5+10x^4+25x^3-28x^2-16}{x-1}=$$

$$=x^6+3x^5+9x^4+19x^3+44x^2+44x+16$$. (*)

Therefore.

(*) not has more real roots, for x>0

Solution: x=1

Regards.

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#### anemone

##### MHB POTW Director
Staff member
Hello.

Brute force.

$$x^4-x^3+x^2+3x+4<(x^2+x+2)^2$$

$$x^4-x^3+x^2+3x+4<x^4+2x^3+5x^2+4x+4$$

$$3x^3+4x+x>0 \rightarrow{}x>0$$

$$x^4-x^3+x^2+3x+4<16$$

$$x^4-x^3+x^2+3x-12<0$$

$$(x^2-3)(x^2-x+4)<0$$

$$x< \sqrt{3}$$

$$\dfrac{1}{(x^2+x+2)^2}+\dfrac{1}{16}-\dfrac{1}{x^4-x^3+x^2+3x+4}=0$$

$$x^8+x^7+4x^6+4x^5+15x^4+23x^3+36x^2+28x+16=16(3x^3+4x^2+x)$$

$$x^8+x^7+4x^6+4x^5+15x^4-25x^3-28x^2+12x+16=0$$

Adding the coefficients, it is "0". Therefore x=1 it's root.

$$\dfrac{x^8+x^7+4x^6+4x^5+15x^4-25x^3-28x^2+12x+16}{x-1}=$$

$$=x^7+2x^6+6x^5+10x^4+25x^3-28x-16$$

Adding the coefficients, it is "0". Therefore x=1 it's root.

$$\dfrac{x^7+2x^6+6x^5+10x^4+25x^3-28x^2-16}{x-1}=$$

$$=x^6+3x^5+9x^4+19x^3+44x^2+44x+16$$. (*)

Therefore.

(*) not has more real roots, for x>0

Solution: x=1

View attachment 2002

Regards.
Yes, the brute force answer is correct and thanks for participating, mente oscura!

My solution:

The given equation can be rewritten as $x^8+x^7+4x^6+4x^5+15x^4-25x^3-28x^2+12x+16=0$, provided $x^4-x^3+x^2+3x+4\ne 0$.

If we let $f(x)=x^8+x^7+4x^6+4x^5+15x^4-25x^3-28x^2+12x+16$ and finding its first and second derivative we have $f'(x)=8x^7+7x^6+24x^5+20x^4+60x^3-75x^2-56x+12$,
$f''(x)=56x^6+42x^5+120x^4+80x^3+180x^2-150x-56$,

and observe that both $f(1)$ and $f'(1)=0$, but $f''(1)\ne 0$, hence we know $x=1$ is a repeated root of is a root of multiplicity 2.

Use polynomial long division to find the remaining factors, we get

\begin{align*}f(x)&=x^8+x^7+4x^6+4x^5+15x^4-25x^3-28x^2+12x+16\\&=(x-1)^2(x^6+3x^5+9x^4+19x^3+44x^2+44x+16)\\&=(x-1)^2((x^2+x+1)^3+3x^4+12x^3+38x^2+38x+15)\\&=(x-1)^2((x^2+x+1)^3+3x^2(x^2+4x+4)+26x^2+38x+15)\\& =(x-1)^2((x^2+x+1)^3+3x^2(x^2+4x+4)+26\left(\left(x+ \dfrac{19}{26} \right)^2+\dfrac{29}{676} \right)\\&=(x-1)^2(\left(\left(x+ \dfrac{1}{2} \right)^2+\dfrac{3}{4} \right)^3+3x^2(x+2)^2+26\left(\left(x+ \dfrac{19}{26} \right)^2+\dfrac{29}{676} \right))\end{align*}

Since the second factor is always greater than zero, it is safe at this point to conclude that the given equation has only one solution, and that is $x=1$.