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Find the sum of the roots, real and non-real, of the equation $$x^{2001}+\left(\frac 12-x\right)^{2001}=0$$, given that there are no multiple roots.

Jun 20, 2019

#1
+25342
+3

Find the sum of the roots, real and non-real, of the equation $$x^{2001}+\left(\dfrac 12-x\right)^{2001}=0$$,
given that there are no multiple roots.

$$\begin{array}{|lcll|} \hline \mathbf{x^{2001}+\left(\dfrac 12-x\right)^{2001}} &=& {0} \\\\ x^{2001} + \dbinom{2001}{0}\left(\dfrac12\right)^{2001} - \dbinom{2001}{1} \left(\dfrac12\right)^{2000}x^1+\ldots + \\ - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999} + \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} - \dbinom{2001}{2001}x^{2001}&=& 0 \\\\ x^{2001} + \dbinom{2001}{0}\left(\dfrac12\right)^{2001} - \dbinom{2001}{1} \left(\dfrac12\right)^{2000}x^1+\ldots + \\ - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999} + \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} - x^{2001}&=& 0 \\\\ \dbinom{2001}{0}\left(\dfrac12\right)^{2001} - \dbinom{2001}{1} \left(\dfrac12\right)^{2000}x^1+\ldots + \\ - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999} + \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} &=& 0 \\ \hline \end{array}$$

$$\begin{array}{|lcll|} \hline \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999}+- \ldots + \dbinom{2001}{0}\left(\dfrac12\right)^{2001} &=& 0 \\\\ 1000.5x^{2000} - 500250 x^{1999}+- \ldots + \left(\dfrac12\right)^{2001} &=& 0 \quad | \quad : 1000.5 \\\\ x^{2000} - \dfrac{500250}{1000.5}x^{1999}+- \ldots + \dfrac{\left(\dfrac12\right)^{2001}}{1000.5} &=& 0 \\\\ x^{2000} - 500x^{1999}+- \ldots + \dfrac{\left(\dfrac12\right)^{2001}}{1000.5} &=& 0 \\\\ x^{2000} \underbrace{- 500}_{=-\sum \limits_{k=1}^{2000} x_k}x^{1999}+- \ldots + \dfrac{\left(\dfrac12\right)^{2001}}{1000.5} &=& 0 \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline -\sum \limits_{k=1}^{2000} x_k &=& -500 \\ \mathbf{\sum \limits_{k=1}^{2000} x_k} &=& \mathbf{500} \\ \hline \end{array}$$

The sum of the roots is 500

Jun 20, 2019

#1
+25342
+3

Find the sum of the roots, real and non-real, of the equation $$x^{2001}+\left(\dfrac 12-x\right)^{2001}=0$$,
given that there are no multiple roots.

$$\begin{array}{|lcll|} \hline \mathbf{x^{2001}+\left(\dfrac 12-x\right)^{2001}} &=& {0} \\\\ x^{2001} + \dbinom{2001}{0}\left(\dfrac12\right)^{2001} - \dbinom{2001}{1} \left(\dfrac12\right)^{2000}x^1+\ldots + \\ - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999} + \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} - \dbinom{2001}{2001}x^{2001}&=& 0 \\\\ x^{2001} + \dbinom{2001}{0}\left(\dfrac12\right)^{2001} - \dbinom{2001}{1} \left(\dfrac12\right)^{2000}x^1+\ldots + \\ - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999} + \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} - x^{2001}&=& 0 \\\\ \dbinom{2001}{0}\left(\dfrac12\right)^{2001} - \dbinom{2001}{1} \left(\dfrac12\right)^{2000}x^1+\ldots + \\ - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999} + \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} &=& 0 \\ \hline \end{array}$$

$$\begin{array}{|lcll|} \hline \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999}+- \ldots + \dbinom{2001}{0}\left(\dfrac12\right)^{2001} &=& 0 \\\\ 1000.5x^{2000} - 500250 x^{1999}+- \ldots + \left(\dfrac12\right)^{2001} &=& 0 \quad | \quad : 1000.5 \\\\ x^{2000} - \dfrac{500250}{1000.5}x^{1999}+- \ldots + \dfrac{\left(\dfrac12\right)^{2001}}{1000.5} &=& 0 \\\\ x^{2000} - 500x^{1999}+- \ldots + \dfrac{\left(\dfrac12\right)^{2001}}{1000.5} &=& 0 \\\\ x^{2000} \underbrace{- 500}_{=-\sum \limits_{k=1}^{2000} x_k}x^{1999}+- \ldots + \dfrac{\left(\dfrac12\right)^{2001}}{1000.5} &=& 0 \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline -\sum \limits_{k=1}^{2000} x_k &=& -500 \\ \mathbf{\sum \limits_{k=1}^{2000} x_k} &=& \mathbf{500} \\ \hline \end{array}$$

The sum of the roots is 500

heureka Jun 20, 2019