Solve Equation II: $\dfrac{1}{(x^2+x+2)^2}$

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In summary, the general form of the equation is $\dfrac{1}{(x^2+x+2)^2} = 0$, and the possible solutions are any real numbers that make the denominator $(x^2+x+2)^2$ equal to 0, which are $x = -1 \pm i$. However, there are no real solutions to this equation as it would result in division by 0. This equation can be solved by setting the numerator equal to 0, but since the numerator is always 1, there are no solutions. The equation cannot be simplified further and the domain is all real numbers except $-1 \pm i$.
  • #1
anemone
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Solve the equation below for real solutions:

$\dfrac{1}{(x^2+x+2)^2}+\dfrac{1}{16}-\dfrac{1}{x^4-x^3+x^2+3x+4}=0$
 
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  • #2
anemone said:
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
 
  • #3
anemone said:
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. (Crying)

[tex]x^4-x^3+x^2+3x+4<(x^2+x+2)^2[/tex]

[tex]x^4-x^3+x^2+3x+4<x^4+2x^3+5x^2+4x+4[/tex]

[tex]3x^3+4x+x>0 \rightarrow{}x>0[/tex][tex]x^4-x^3+x^2+3x+4<16[/tex]

[tex]x^4-x^3+x^2+3x-12<0[/tex]

[tex](x^2-3)(x^2-x+4)<0[/tex]

[tex]x< \sqrt{3}[/tex][tex]\dfrac{1}{(x^2+x+2)^2}+\dfrac{1}{16}-\dfrac{1}{x^4-x^3+x^2+3x+4}=0[/tex]

[tex]x^8+x^7+4x^6+4x^5+15x^4+23x^3+36x^2+28x+16=16(3x^3+4x^2+x)[/tex]

[tex]x^8+x^7+4x^6+4x^5+15x^4-25x^3-28x^2+12x+16=0[/tex]

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

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

[tex]=x^7+2x^6+6x^5+10x^4+25x^3-28x-16[/tex]

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

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

[tex]=x^6+3x^5+9x^4+19x^3+44x^2+44x+16[/tex]. (*)

Therefore.

(*) not has more real roots, for x>0Solution: x=1
(Sleepy)

View attachment 2002

Regards.
 

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  • #4
mente oscura said:
Hello.

Brute force. (Crying)

[tex]x^4-x^3+x^2+3x+4<(x^2+x+2)^2[/tex]

[tex]x^4-x^3+x^2+3x+4<x^4+2x^3+5x^2+4x+4[/tex]

[tex]3x^3+4x+x>0 \rightarrow{}x>0[/tex][tex]x^4-x^3+x^2+3x+4<16[/tex]

[tex]x^4-x^3+x^2+3x-12<0[/tex]

[tex](x^2-3)(x^2-x+4)<0[/tex]

[tex]x< \sqrt{3}[/tex][tex]\dfrac{1}{(x^2+x+2)^2}+\dfrac{1}{16}-\dfrac{1}{x^4-x^3+x^2+3x+4}=0[/tex]

[tex]x^8+x^7+4x^6+4x^5+15x^4+23x^3+36x^2+28x+16=16(3x^3+4x^2+x)[/tex]

[tex]x^8+x^7+4x^6+4x^5+15x^4-25x^3-28x^2+12x+16=0[/tex]

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

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

[tex]=x^7+2x^6+6x^5+10x^4+25x^3-28x-16[/tex]

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

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

[tex]=x^6+3x^5+9x^4+19x^3+44x^2+44x+16[/tex]. (*)

Therefore.

(*) not has more real roots, for x>0Solution: x=1
(Sleepy)

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 $x=1$.:)
 
  • #5


To solve this equation, we can first combine the fractions on the left side by finding a common denominator. This will give us:

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

Next, we can simplify the numerator by expanding the squared terms and combining like terms. This will give us:

$\dfrac{x^4+8x^3+33x^2+64x+68}{(x^2+x+2)^2}=0$

From here, we can factor the numerator to get:

$\dfrac{(x+2)(x+4)(x^2+3x+17)}{(x^2+x+2)^2}=0$

Since the numerator is now factored, we can set each factor equal to zero and solve for x. This gives us the following solutions:

$x=-2, x=-4, x=\dfrac{-3\pm\sqrt{61}i}{2}$

Therefore, the real solutions to this equation are $x=-2$ and $x=-4$. The other two solutions involve imaginary numbers and are not considered real solutions.
 

FAQ: Solve Equation II: $\dfrac{1}{(x^2+x+2)^2}$

What is the general form of the equation?

The general form of the equation is $\dfrac{1}{(x^2+x+2)^2} = 0$.

What are the possible solutions of the equation?

The possible solutions of the equation are any real numbers that make the denominator $(x^2+x+2)^2$ equal to 0, which are $x = -1 \pm i$. However, since the equation is in the form of $\dfrac{1}{(x^2+x+2)^2}$, the value of $x$ cannot be equal to $-1 \pm i$ as it would result in division by 0. Therefore, there are no real solutions to this equation.

How do you solve this equation?

This equation can be solved by setting the numerator equal to 0 and solving for $x$. However, since the numerator is always equal to 1, there are no values of $x$ that can satisfy this equation. Therefore, this equation has no solutions.

Can this equation be simplified?

No, this equation cannot be simplified further as it is already in its simplest form.

What is the domain of this equation?

The domain of this equation is all real numbers except $-1 \pm i$, as these values would result in division by 0. Therefore, the domain is $(-\infty, -1) \cup (-1, \infty)$.

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