# Find the sum of the roots of the equation?

What is the sum of the roots of the equation

$$(x − 1) + (x − 2)^2 + (x − 3)^3 + ... + (x − 10)^{10} = 0$$?

When i expand this equation, it become in the power of 10 and its get complicated. Now what i am thinking is the sum of roots will be equal to the sum of coefficents of x^9 .So i just need to evaluate coefficent of x^9 in the term $$(x-10)x^{10}$$. Am in right in thinking?

But is there is any other easier way by which i can calculate?

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Yes, you are right. Take a look at: en.wikipedia.org/wiki/Vieta's_formulas –  Beni Bogosel Apr 16 '12 at 11:46
I would think you'd be after the coeff of $x^9$ in $(x-9)^9+(x-10)^{10}$. –  Gerry Myerson Apr 16 '12 at 11:48
Ya,I dint mention it as the coefficent of x^9 in (x-9)^9 will be 1 for sure. –  vikiiii Apr 16 '12 at 11:50

Coefficient of $x^9$ is $$1+ {10 \choose 1} (-10) = 1 - 100 = -99$$

The sum of the roots is therefore $99$ Do you know why?

I am adding this explanation on request: Although, I would recommend you have to read more on Vieta's formula.

If $\alpha$ and $\beta$ are roots of a quadratic equation, then $(x-\alpha)(x-\beta) = x^2-(\alpha+\beta)x+\alpha \beta$ and the absolute value of the coefficient of $x$ is the sum of of the roots. (In this case $\alpha+\beta$.

Similarly a tenth degree polynomial, say has roots $\alpha_1, \alpha_2, \dots \alpha_{10}$ , then the polynomial

$$(x-\alpha_1)(x-\alpha_2)\dots(x-\alpha_{10}) = x^{10}-(\sum_{i=1}^{10} \alpha_i)x^9 +\dots +\prod_{i=1}^{10} \alpha_i$$

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Its right but Can you please explain? –  vikiiii Apr 16 '12 at 11:54
@vikiiii I added explanation as you requested. –  Kirthi Raman Apr 16 '12 at 12:10
Raman But here we dont know the roots of equation. So how to find their sum. –  vikiiii Apr 23 '12 at 2:38
@vikiii the sum of the roots is the absolute value of the coefficient of $x^9$ which is the absolute value of $1+ {10 \choose 1} (-10) = 1 - 100 = -99$ and that absolute value is $99$. For instance what is the sum of the roots of $(x-2)^2+(x-4)^2=0$? which is $2x^2-12x+20=2(x^2-6x+10)=0$. The sum of the roots is the absolute value of the coefficient of $x$ and that is 6. –  Kirthi Raman Apr 23 '12 at 11:28

Hint $\:$ Shift $\rm\: x = y+10,\:$ so $\rm\:y^{10} + (y+1)^9 +\cdots\: = y^{10} + y^9 + \cdots\:$ has root sum $\:\!-1,\:$ so $\rm\:x_1+\cdots + x_{10} =\: (y_1\!+\!10)+\cdots+(y_{10}\!+\!10) =\: y_1+\cdots+y_{10}+100 = -1 + 100 = 99.$

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Can you explain a little bit more? –  vikiiii Apr 23 '12 at 2:38
@vikiiii What is proving problematic? –  Bill Dubuque Apr 23 '12 at 2:45