+0

0
64
2

Find the number of real roots of 2x^2001 + 3x^2000 + 2x^1999 + 3x^1998 + ...+ 2x + 3 = 0.

Jan 23, 2020

#1
0

The equation has no real roots, so the answer is 0.

Jan 23, 2020
#2
+24097
+1

Find the number of real roots of

$$2x^{2001} + 3x^{2000} + 2x^{1999} + 3x^{1998} + \ldots + 2x + 3 = 0$$.

I assume:

$$\begin{array}{|rcll|} \hline 2x^{2001} + 3x^{2000} + 2x^{1999} + 3x^{1998} + \ldots + 2x + 3 &=& 0 \\ 3+2x+3x^2+2x^3+\ldots + 3x^{1998}+ 2x^{1999}+ 3x^{2000}+2x^{2001}&=& 0 \\ (3+2x)\left(1+x^2+x^4+x^6+\ldots +x^{1998}+x^{2000} \right) &=& 0 \\\\ \boxed{1+x^2+x^4+x^6+\ldots +x^{1998}+x^{2000} =\dfrac{1-x^{2002}}{^-x^2}} \\\\ (3+2x)\left( \dfrac{1-x^{2002}}{1-x^2} \right) &=& 0 \\\\ (3+2x)\left( \underbrace{\dfrac{1-x^{2002}}{1-x^2}}_{x\neq \pm 1} \right) &=& 0 \\ 3+2x &=& 0 \\ 2x &=& -3\\\\ \mathbf{x} &=& \mathbf{-\dfrac{3}{2}} \\ \hline \end{array}$$

The  real root  is $$\mathbf{x=-\dfrac{3}{2}}$$

Jan 23, 2020
edited by heureka  Jan 24, 2020