Very Simple Ji........!

The equation can be factored as:

$(2x+3)(1+x^2+x^4+x^6+...+x^{2000})=0$

0nly one factor gives a real root.

It is an equation in one variable, simple. . .

2x^2001+3x^2000+.............................. + 2x+3=0 x^2000(2x+3)+x^1998(2x+3)+........................+ 1(2x+3)=0 (x^2000+x^1998+............................+x^2+1)(2x+3)=0 ONE ROOT : x=-3/2 x^2000+x^1998+.......................+ x^2=-1 Sum of positive integers is always greater than or equal to zero. therefore only one real root.

Grouping the even and odd exponents:

$2\sum\limits_{i=0}^{1000}{x^{2i+1}} + 3\sum\limits_{i=0}^{1000}{x^{2i}} = 0$

Because $x^{a+b} = x^ax^b$ , an $x$ is factored from the odd exponents:

$2x\sum\limits_{i=0}^{1000}{x^{2i}} = -3\sum\limits_{i=0}^{1000}{x^{2i}}$

Because $x = 0$ is not a solution to the original equation, $\sum{x^{2i}}$ is non-zero. Dividing each side by it yields:

$2x = -3$

So $x = -3/2$ , the only root.