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September 29th, 2018, 08:05 AM   #1
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An inequality

Good evening to all,

Solve the inequality $\displaystyle ||x^2+2ix+3|'+4|+5x^2<0$.

All the best,

Integrator
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September 29th, 2018, 10:47 AM   #2
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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 $
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September 29th, 2018, 10:24 PM   #3
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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.
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September 30th, 2018, 06:07 AM   #4
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Does x have to be real?
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September 30th, 2018, 06:37 AM   #5
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Quote:
Originally Posted by skipjack View Post
Does x have to be real?
Hello,

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

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Integrator
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October 1st, 2018, 09:40 PM   #6
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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 $\displaystyle u,v\in \mathbb R$?

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Integrator

Last edited by Integrator; October 1st, 2018 at 09:48 PM.
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