1

$nx^{(n+1)}-(n+1)x^n+1=0$

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

doraemonpaul
  • 16,178
  • 3
  • 31
  • 75
Florin M.
  • 635

3 Answers3

3

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 $1$.

UrošSlovenija
  • 915
  • Thanks again! And the equation : $nx^{n-1} + (n-1)x^{n-2}+\dots+x+1=0$ has no solutions ? – Florin M. May 29 '13 at 06:39
  • 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. – UrošSlovenija May 29 '13 at 07:42
  • 1
    Or differentiate, getting $n (n + 1) (x^n - x^{n - 1})$, which has the zero 1 in common with the original. – vonbrand May 29 '13 at 12:11
1

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

Alex Wertheim
  • 20,278
0

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

$f'(x)=n(n+1)x^n-n(n+1)x^{n-1}=n(n+1)x^{n-1}(x-1)$

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 $x=1$

Abhra Abir Kundu
  • 8,150