Hello!

Solve the inequality , where .

Thank You!

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- May 10th 2013, 09:33 PMDacuAn inequality
Hello!

Solve the inequality , where .

Thank You! - May 10th 2013, 10:00 PMProve ItRe: An inequality
First, you have an inequality involving complex coefficients, but you are told it has to be less than a real number. Since complex numbers can NOT be ordered, you require putting in values which will give you a REAL solution.

So that means that the expression needs to be real. So if we let where , then that gives

For this expression to be real, that means

Case 1: would give

We require this expression to be , so we have

Case 2: would give

We require this expression to be , so we have , but since there are not any solutions.

So that means that the solution to your equation is where . - May 13th 2013, 01:51 AMibduttRe: An inequality
Dear Prove it,

can we proceed like this: Let us put the equation equal to 0

3x^2 + 4ix + 5 = 0 OR

- 3 ( -ix ) ^ 2 + 4 ( ix ) + 5 = 0 Let ix = a We get

3a^2 - 4a - 5 = 0, By using quadratic formula we get

a = [ 4 =/- 4 sqrt ( 19 )] / 6 =[ 2 +/- 2 sqrt ( 19 )] / 3

That means ix = [ 2 +/- 2 sqrt ( 19 )] / 3 - May 13th 2013, 01:54 AMProve ItRe: An inequality
Well that gives you where the function IS 0, but that's not what the question was asking. But it's a start :)