complex number question (1 Viewer)

[Stefano]

KFunk : Haha - funny!

acmilan: Don't you think my way is faster though ?

 

[acmilan]

For a quadratic, yes

 

[Stefano]

For a quadratic, yes

Yup!

[Your method is obviously more formal and flexible, though]

 

[pLuvia]

@ = {-b +/- root.[b²-4ac]}/2a (Quadratic Formula)

Therefore, the roots are:

1. {-b + root.[b²-4ac]}/2a (@)

2. {-b - root.[b²-4ac]}/2a (@-bar)


Understood?

ah ok thanks mate

 

[pLuvia]

Complex conjugate pairs... (Do you need a proof? Because if you do I don't have one )

But can I write that? And yeh I need some proof

 

[Stefano]

ah ok thanks mate

No problem mate.

 

[acmilan]

But can I write that? And yeh I need some proof

afaik, you can quote that complex roots come in conjugate pairs without proof (correct me if im wrong, i didnt do 4 unit). But if you want proof, my post above shows how to do it, as did KFunk's.

 

[Stefano]

For this particular question you need some proof.

For longer (and sorry to say it, but difficult) questions you can just quote it without proof before moving onto the next part.