There's nothing told about $n$, I guess $ n \in N $. I would like any kind explanations, thanks! I appreciate your time.

  • $\begingroup$ Guys, sorry. The first x is at power : n+1 . I can't edit it. Someone help me. THanks $\endgroup$ – Florin M. May 29 '13 at 5:59

Hint: $ x =1$ solves the equation. Factoring gives $$ (x-1)(nx^n - x^{n-1}-x^{n-2}- \dots -x -1) = 0 $$ and you can see $x=1$ is again a zero of the second factor. Factoring further $$ (x-1)^2(nx^{n-1} + (n-1)x^{n-2}+\dots+2x+1) = 0 $$ and you can conclude that for $n=1,2,\dots$ the equation has a double zero at $1$.

  • $\begingroup$ Thanks again! And the equation : $nx^{n-1} + (n-1)x^{n-2}+\dots+x+1=0$ has no solutions ? $\endgroup$ – Florin M. May 29 '13 at 6:39
  • $\begingroup$ I guess that for odd $n$ there is another real solution. For even $n$ there are only solution in complex plane with nonzero imaginary part. $\endgroup$ – UrošSlovenija May 29 '13 at 7:42
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    $\begingroup$ Or differentiate, getting $n (n + 1) (x^n - x^{n - 1})$, which has the zero 1 in common with the original. $\endgroup$ – vonbrand May 29 '13 at 12:11

Hint: if you distribute the equation, you get the expression $2 - x^{n} = 0$.

  • $\begingroup$ You probably saw the version that looked like the exponent $n+1$ was a factor. $\endgroup$ – Ross Millikan Jun 6 '13 at 0:23

Let $f(x)=nx^{(n+1)}-(n+1)x^n+1$ $n\in N,x>0$


So we have $f'(x)>0 \forall x>1$

and we have $f'(x)<0\forall 0<x<0$

This implies that the function is increasing for all $x>1$ and decreasing for $0<x<1$

Now $f(1)=0$ as $f(x)$ is increasing for $x>1$ so we have $f(x)>f(1)=0,\forall x>1$

And as it is decresing for $0<x<1$ so we have $f(x)<f(1)$ for $0<x<1$.

So the only (positive)solution of this equation is $x=1$

  • $\begingroup$ Thanks, dear Abhra! $f'(x)<0\forall 0<x<0$, is from $0<x<1$, right? $\endgroup$ – Florin M. May 29 '13 at 6:55
  • $\begingroup$ yes you are right $\endgroup$ – Abhra Abir Kundu May 29 '13 at 6:57

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