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Factoring a complex number

Spencer278

Diamond Member
Can anyone tell me how to factor a formula into complex conjate roots? I tried google but it seems no one has written anything useful.
 
Originally posted by: Spencer278
Can anyone tell me how to factor a formula into complex conjate roots? I tried google but it seems no one has written anything useful.

Try using conjugate instead of "conjate" and you will find something in google

Regards

ng
 
Originally posted by: ngvepforever2
Originally posted by: Spencer278
Can anyone tell me how to factor a formula into complex conjate roots? I tried google but it seems no one has written anything useful.

Try using conjugate instead of "conjate" and you will find something in google

Regards

ng

Thanks but you can take your thread crap in another thread please.
 
Factoring: x^2 + 4x + 6

a = 1, b = 4, c = 6

So, (-4 +- sqrt (4^2 - 4*1*6)) / (2*1)

Simplifies to: (-4 +- sqrt(16 - 24)) / 2

-> -2 +- sqrt(-8)/2
-> -2 +- i*sqrt(8)/2

So the two solutions are -2 + i*sqrt(8)/2 and -2-i*sqrt(8)/2
 
Originally posted by: Spencer278
Originally posted by: ngvepforever2
Originally posted by: Spencer278
Can anyone tell me how to factor a formula into complex conjate roots? I tried google but it seems no one has written anything useful.

Try using conjugate instead of "conjate" and you will find something in google

Regards

ng

Thanks but you can take your thread crap in another thread please.

hey fvckhead, that wasnt a thread crap. he was telling you how to spell it correctly, you dumb sh!t.
 
Either way, all you need to know is that i (or j if you're in Electrical or Computer Engineering) = sqrt(-1)

Therefore i^2 = 1

Then just simplify.

--Mark
 
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