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Find the number of real roots of $2x^{2001} + 3x^{2000} + 2x^{1999} + 3x^{1998} + \dots + 2x + 3 = 0.$

Guest Apr 12, 2020

Guest · Answer 1 · 2020-04-13T03:48:35

There are 5 real roots.

MaxWong · Answer 2 · 2020-04-14T10:30:35

$2x^{2001} + 3x^{2000} + 2x^{1999} + 3x^{1998} + \cdots + 2x + 3 = 0\\ (2x + 3)(x^{2020} + x^{2018} + \cdots + 1) = 0$

Because $x^{2020} + x^{2018} + \cdots + 1 > 0$ , the only possibility is $2x + 3 = 0$ .

Solving gives $x = -\dfrac32$ .
There is only 1 real root.