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Solve the equation $$\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0.$$ For $x\ne -4;-3;-2;-1;0$ we have $$(x+1)(x+2)(x+3)(x+4)+x(x+2)(x+3)(x+4)+x(x+1)(x+3)(x+4)+\text{...}=0$$ Most likely that's not the author's intention. I have tried to substitute $t=x+2$ to get $$\dfrac{1}{t-2}+\dfrac{1}{t-1}+\dfrac{1}{t}+\dfrac{1}{t+1}+\dfrac{1}{t+2}=0$$ which actually isn't easier to work with than the original problem.

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  • $\begingroup$ In any case, you will have a fourth order polynomial equation in $x$, and there are four distinct roots. $\endgroup$
    – mjw
    Commented Apr 28, 2021 at 21:02
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    $\begingroup$ I would actually argue the second polynomial is easier to work with! You can pull the same trick with the original problem of multiplying by $(t-2)(t-1)...$ to arrive at the wonderfully simple polynomial $5t^4-15t^2+4$ (thanks Wolfram|Alpha) $\endgroup$
    – Andrew L
    Commented Apr 28, 2021 at 21:05

6 Answers 6

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Let $y=x+2$. Then your equation becomes$$\frac1{y-2}+\frac1{y-1}+\frac1y+\frac1{y+1}+\frac1{y+2}=0$$or$$\frac{5 y^4-15 y^2+4}{(y-2) (y-1) y (y+1) (y+2)}=0,$$which can be solved using the quadratic formula. Its roots are $\pm\sqrt{\frac{15\pm\sqrt{145}}{10}}$, and therefore the roots of the original equation are $-2\pm\sqrt{\frac{15\pm\sqrt{145}}{10}}$.

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Perhaps this will help: $${1\over x}+{1\over x+4} = {2x+4\over x^2+4x}$$ $${1\over x+1}+{1\over x+3} = {2x+4\over x^2+4x+3}$$

$${1\over x+2} = {x+2\over x^2+4x+4}$$

and now we have $$ {x+2\over x^2+4x+4}+ {2x+4\over x^2+4x} + {2x+4\over x^2+4x+3}=0$$

Clearly $x\ne -2$ and let $t=x^2+4x$ then you have:

$${1\over t+4}+{2\over t+3} + {2\over t}=0$$ i.e. $$5t^2 + 25t + 24=0$$

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  • $\begingroup$ That's actually not easier to work with than my equation after substituting $t=x+2$, because the roots of $$5t^2+25t+24=0$$ are $x_{1,2}=\dfrac{-25\pm\sqrt{145}}{10}=-\dfrac{5}{2}\pm\sqrt{\dfrac{145}{100}}=-\dfrac{5}{2}\pm\dfrac{1}{2}\sqrt{\dfrac{29}{5}}$. So we have $$x^2+4x=-\dfrac{5}{2}+\dfrac{1}{2}\sqrt{\dfrac{29}{5}}$$ which I cannot solve. $\endgroup$ Commented Apr 29, 2021 at 18:51
  • $\begingroup$ $$x^2+4x +\color{red}4 = {3\over 2} + {1\over 2}\sqrt{29\over 5}$$ so $$x+2 = \pm\sqrt{{3\over 2} + {1\over 2}\sqrt{29\over 5}}$$ $\endgroup$
    – nonuser
    Commented Apr 29, 2021 at 18:55
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If $t$ is a solution, then so is $-t$, hence it is worth tring to write things in terms of $u:=t^2$. Note that $$\frac1{t-2}+\frac1{t+2}=\frac{(t+2)+(t-2)}{(t-2)(t+2)}=\frac{2t}{u-4} ,$$ $$\frac1{t-1}+\frac1{t+1}=\frac{(t+1)+(t-1)}{(t-1)(t+1)}=\frac{2t}{u-1} ,$$ and of course $$\frac1t=\frac{2t}{2u} $$ so that we want to solve $$\frac1{u-4}+\frac1{u-1}+\frac1{2u}=0 $$ This gives you only a quadratic numerator.

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  • $\begingroup$ Thank you for the response! May I ask you why if $t$ is a solution, then so is $-t$? $\endgroup$ Commented Apr 28, 2021 at 21:30
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    $\begingroup$ Inserting $ \ -t \ $ into $$\dfrac{1}{t-2}+\dfrac{1}{t-1}+\dfrac{1}{t}+\dfrac{1}{t+1}+\dfrac{1}{t+2}=0$$ makes the left side of the equation exactly its negative, which does not change the solutions. $\endgroup$
    – user882145
    Commented Apr 29, 2021 at 2:49
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You can write it as:

$$\sum_{j=0}^4\prod_{\begin{matrix}i=0\\i\ne j\end{matrix}}^4(x+i)=0$$ which might make it easier to type into an equation expander. Otherwise, note that expanded it gives: $$5x^4+40x^3+105x^2+100x+24=0$$ (I believe anyway I did not check this by hand). This does in fact have four real roots.

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Let $f$ be the conveniently translated function defined by :

$$f(t):=\dfrac{1}{t-2}+\dfrac{1}{t-1}+\dfrac{1}{t}+\dfrac{1}{t+1}+\dfrac{1}{t+2}$$

What I would like to underline is that a graphical representation provides a good insight for such questions. In particular, it can help to anticipate (or verify) a certain number of results, in particular the fact that the roots of $f(t)=0$ are all real, separated by integers $-2,-1,0,1,2$ and symmetric with respect to $0$ due to the fact that $f$ is an odd function.

enter image description here

Indeed, this function has a decreasing behavior (the derivative $f'$ is clearly negative) from $+\infty$ to $+\infty$ on each of the intervals: $[k,k+1], \ k=-2,-1,0,1$ delimitated by vertical asymptotes ; the two external branches do not provide any supplementary root.

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Most likely that's not the author's intention

It could very well be. If we replace $y = x + 2$ we have

$\frac 1{y-2} + \frac 1{y-1} +\frac 1y + \frac 1{y+1}+\frac 1{y+2} = 0$ so

$(y-1)y(y+1)(y+2) + (y-2)y(y+1)(y+2) + (y-2)(y-1)(y+1)(y+2) + (y-2)(y-1)y(y+2) + (y-2)(y-1)y(y-1) =0$

$(y^2-1)(y^2+2y) + (y^2-4)(y^2+ y) + (y^2-4)(y^2 - 1) + (y^2 -4)(y^2-y) + (y^2-1)(y^2 - y)=0$

$(y^2-1)[(y^2 + 2y)+(y^2 - 2y)] + (y^2-4)[(y^2 + y)+(y^2-y)] + (y^2 -4)(y^2 -1) = 0$

$(y^2-1)2y^2 + (y^2-4)2y^2 + (y^2 -4)(y^2 -1) = 0$

$2y^2(2y^2 - 5) + (y^4 -5y^2 +4) = 0$

$5y^4 - 15y^2 + 4 = 0$

$y^2 = \frac {15 \pm \sqrt{15^2-80}}{10}= \frac {15\pm \sqrt {145}}{10}$

$y = \pm\frac {15\pm \sqrt {145}}{10}$

$x = \pm \frac {15\pm \sqrt {145}}{10}-2$

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