September 29th, 2018, 08:05 AM  #1 
Newbie Joined: Aug 2018 From: RomÃ¢nia Posts: 24 Thanks: 2  An inequality
Good evening to all, Solve the inequality $\displaystyle x^2+2ix+3'+4+5x^2<0$. All the best, Integrator 
September 29th, 2018, 10:47 AM  #2 
Newbie Joined: Sep 2018 From: Poland Posts: 1 Thanks: 0  Inequality
$ x^2 +2ix + 3 ' +4 +5x^2 <0 $ $  (x^2 +2ix + i^2 +4)' +4  +5x^2<0$ $ [(x +i)^2 +4]' +4  +5x^2<0 $ $ 2(x +i) + 4  +5x^2 <0 $ $ 2x+2 +i +5x^2 <0 $ $ 2\sqrt{(x+2)^2 +1^2} +5x^2 <0 $ $ x\in \emptyset $ 
September 29th, 2018, 10:24 PM  #3 
Newbie Joined: Aug 2018 From: RomÃ¢nia Posts: 24 Thanks: 2 
Good morning to all, Thousands of apologies!I reformulate the statement of the problem: Solve the inequality $\displaystyle x^2+2ix+3'+4+5x^2<0$ where $\displaystyle i^2=1$. All the best, Integrator Last edited by Integrator; September 29th, 2018 at 10:37 PM. 
September 30th, 2018, 06:07 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,269 Thanks: 1958 
Does x have to be real?

September 30th, 2018, 06:37 AM  #5 
Newbie Joined: Aug 2018 From: RomÃ¢nia Posts: 24 Thanks: 2  
October 1st, 2018, 09:40 PM  #6 
Newbie Joined: Aug 2018 From: RomÃ¢nia Posts: 24 Thanks: 2 
Good morning to all, Some say there are solutions...How can we solve this inequality?What are these solutions? What resulting , if we consider $\displaystyle x=u+vi$ where $\displaystyle i^2=1$ and $\displaystyle u,v\in \mathbb R$? All the best, Integrator Last edited by Integrator; October 1st, 2018 at 09:48 PM. 

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