+0  
 
0
236
3
avatar

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

 May 12, 2019
 #1
avatar+111456 
+2

2x^2001 + 3x^2000 + 2x^1999 + 3x^1998 + .....+ 2x + 3    note that we can write

 

x^2000 ( 2x + 3) + x^1998(2x + 3) + x^1996(2x + 3) +  ....+ x^2(2x + 3) + x^0 (2x + 3) =

 

x^2000(2x+ 3) +x^!998(2x + 3) + x^1996(2x + 3) + ......+ x^2(2x + 3)  + 1(2x + 3)  =

 

(2x + 3) ( x^2000 + x^1998 + x^1996 + .....+ x^2 + 1)

 

Since the second polynomial is > 0  for all x  the only real root  is

 

2x + 3  =0

2x = - 3

x = -3/2

 

 

cool cool cool

 May 12, 2019
 #2
avatar+6196 
+1

very clever dr. Jones!

Rom  May 12, 2019
 #3
avatar+111456 
0

THX, Rom  !!!

 

cool cool cool

CPhill  May 12, 2019

49 Online Users

avatar
avatar