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 September 29th, 2018, 08:05 AM #1 Newbie   Joined: Aug 2018 From: RomÃ¢nia Posts: 17 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 $||x^2 +2ix + 3 |' +4| +5x^2 <0$ $| (x^2 +2ix + i^2 +4)' +4 | +5x^2<0$ $|[(x +i)^2 +4]' +4 | +5x^2<0$ $|2(x +i) + 4 | +5x^2 <0$ $2|x+2 +i| +5x^2 <0$ $2\sqrt{(x+2)^2 +1^2} +5x^2 <0$ $x\in \emptyset$
 September 29th, 2018, 10:24 PM #3 Newbie   Joined: Aug 2018 From: RomÃ¢nia Posts: 17 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: 19,974 Thanks: 1850 Does x have to be real?
September 30th, 2018, 06:37 AM   #5
Newbie

Joined: Aug 2018
From: RomÃ¢nia

Posts: 17
Thanks: 2

Quote:
 Originally Posted by skipjack Does x have to be real?
Hello,

From solving the inequality will result the nature of $\displaystyle x$.

All the best,

Integrator

 October 1st, 2018, 09:40 PM #6 Newbie   Joined: Aug 2018 From: RomÃ¢nia Posts: 17 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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