Solve the inequality , where .
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 .
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