# Thread: Find all the real numbers m so that..

1. ## Find all the real numbers m so that..

Can any one help me with this:

Find all the REAL numbers m so that the equation

z^3 + (3+i)z^2 -3z - (m+i) = 0

has at least one REAL root.

NOTE: i = (-1)^1/2

2. The equation can be written as
$z^3+3z^2-3z-m+i(z^2-1)=0$
If z is a real root then
$\left\{\begin{array}{ll}z^3+3z^2-3z-m=0\\z^2-1=0\end{array}\right.$
From the second equation we have $z=\pm 1$.
Plugging z in the first equation we get $m=1, \ m=5$

3. Originally Posted by red_dog
The equation can be written as
$z^3+3z^2-3z-m+i(z^2-1)=0$
If z is a real root then
$\left\{\begin{array}{ll}z^3+3z^2-3z-m=0\\z^2-1=0\end{array}\right.$
From the second equation we have $z=\pm 1$.
Plugging z in the first equation we get $m=1, \ m=5$
that was a lot easier than i thought it would be. you don't want to know the things than were going through my head!

4. 10x

5. Originally Posted by N1CKYYY
10x
What is this supposed to mean?

-Dan

6. Originally Posted by topsquark
What is this supposed to mean?

-Dan
inappropriate personal remark removed

7. Originally Posted by N1CKYYY
inappropriate personal remark removed
Why don't you learn some manners and show respect to our senior members?

8. I think this thread has run its course.