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April 5th, 2014, 08:17 AM   #1
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An inequality

To solve the inequality $\displaystyle 3ix^2+4x-5i\leq 0$ where $\displaystyle i^2=-1$.
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April 5th, 2014, 08:52 AM   #2
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$\displaystyle x\,=\,-\sqrt{\frac53}$.
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April 5th, 2014, 06:35 PM   #3
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I couldn't disagree more.

To me the inequality is meaningless because the complex plane has no ordering.

Should there be some modulus symbols somewhere? Is $x$ supposed to be real or complex?
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April 5th, 2014, 07:03 PM   #4
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Try subbing in Olinguito's answer.
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April 5th, 2014, 08:26 PM   #5
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I know where that came from, and yes, by eliminating the imaginary part of the equation it makes some sort of sense of the inequality. But I would suggest that it's a poorly set question if that is the answer that is being sought.
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April 5th, 2014, 09:26 PM   #6
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Perhaps it was a trick question.
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April 5th, 2014, 10:15 PM   #7
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Quote:
Originally Posted by Olinguito View Post
$\displaystyle x\,=\,-\sqrt{\frac53}$.
There are many more values of $\displaystyle x$ that verifies the inequality.
They are,therefore, all values of $\displaystyle x$ that verifies the inequality?
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April 5th, 2014, 11:57 PM   #8
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If you allow $x$ to be complex-valued, you could also have
$$x\,=\,\frac{\pm\sqrt{11}+2i}3$$
By the way, was it meant to be a trick question – i.e. was it meant to provoke the reaction “this inquality is meaningless”? If it was, it was perfect! Had you posted it a few days earlier it would have even made an excellent April fool’s joke.

PS: I’d like to point out that just because an inequality contains complex numbers does not mean that it must be meaningless. For example, the interior of unit circle can be described by $\displaystyle z\overline z<1$. If this inequality makes sense, despite the presence of complex variables, I don’t see why we should not try to make sense of the OP’s inequality.

Last edited by Olinguito; April 6th, 2014 at 12:29 AM.
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April 6th, 2014, 12:47 AM   #9
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Olinguito is correct because 3ix² - 5i = 0 and 4x ≤ 0.
The key to solving this problem correctly is to see how to get rid of the imaginary numbers to solve such problem.
There could be more solutions but I won't go through all of them.
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April 6th, 2014, 06:09 AM   #10
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Quote:
Originally Posted by Olinguito View Post
If you allow $x$ to be complex-valued, you could also have
$$x\,=\,\frac{\pm\sqrt{11}+2i}3$$
Solving of the inequality $\displaystyle 3ix^2+4x-5i\leq 0$ where $\displaystyle i^2=-1$ is simple if it is considered an equivalent equation $\displaystyle 3ix^2+4x-5i=a$ where $\displaystyle i^2=-1$ and $\displaystyle a\in R,a\leq 0$.
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