How did they get the complex roots?

thisonedude

New member
Joined
Sep 11, 2014
Messages
3
So in the answer, they found the complex roots as being 1-2j and 1+2j however, when i did the quadratic formula as being -1+8j and -1-8j, Why is theirs right and mine not?
 

Attachments

  • 1.JPG
    1.JPG
    18.6 KB · Views: 17
So in the answer, they found the complex roots as being 1-2j and 1+2j however, when i did the quadratic formula as being -1+8j and -1-8j, Why is theirs right and mine not?

What are you trying to find the complex roots of? The two roots 1\(\displaystyle \pm\)2j are roots to
\(\displaystyle (s-1)^2 + 4 = 0\).
The two roots -1\(\displaystyle \pm\)8j are roots to
\(\displaystyle (s+1)^2 + 64 = 0\).

Neither of those polynomial expressions appear in your formulas. Of course, if the two roots were -1\(\displaystyle \pm\)2j it would be a different story.
 
So in the answer, they found the complex roots as being 1-2j and 1+2j however, when i did the quadratic formula as being -1+8j and -1-8j, Why is theirs right and mine not?

Short answer - you must have made a mistake. Since we cannot see your work - we cannot tell what/where you made the mistake.

Please show all the steps of your solution - and we might be able to catch your misstep.
 
What are you trying to find the complex roots of? The two roots 1\(\displaystyle \pm\)2j are roots to
\(\displaystyle (s-1)^2 + 4 = 0\).
The two roots -1\(\displaystyle \pm\)8j are roots to
\(\displaystyle (s+1)^2 + 64 = 0\).

Neither of those polynomial expressions appear in your formulas. Of course, if the two roots were -1\(\displaystyle \pm\)2j it would be a different story.

I'm trying to find the roots of s2 ​+2s + 5
 
I'm trying to find the roots of s2 ​+2s + 5

You can use the quadratic formula or note that, as written in the original post that
\(\displaystyle s^2 + 2 s + 5 = (s+1)^2 + 4 \)
and if you want the roots of the equation you have
\(\displaystyle s^2 + 2 s + 5 = (s+1)^2 + 4 = 0\)
=> \(\displaystyle (s+1)^2 = -4 \)
=> \(\displaystyle (s+1) = \pm 2j \)
or
\(\displaystyle s = -1 \pm 2j \)

The quadratic formula gives the same result:
\(\displaystyle s=\frac{-2\pm\sqrt{2^2-4*1*5}}{2}=\frac{-2\pm\sqrt{-16}}{2}=\frac{-2\pm4j}{2}= -1 \pm 2j\)
so it appears that you made a mistake somewhere (did you forget to take the square root of 16?).
 
Last edited:
Top