I am not getting the final part of my partial fraction

[swift orbit]

I really am stuck.
I need 4 equations for 4 unknown terms, but I am only getting 3 after doing partial fractions

[swift orbit]

<@&286206848099549185>

[thin venture]

It's just B/x²

not Bx+C, x² has no complex roots

How is k defined?

[swift orbit]

So. If it was x²+R then the top will be Bx+C?

K is a real number

[thin venture]

Assuming R > 0

[swift orbit]

Please explain

[thin venture]

Hmm first time encountering something like this

[swift orbit]

Wanna see the question?

[thin venture]

ye

[thin venture]

ah i am lost

it's x-k

no x^2-k

[swift orbit]

Can you explain why R < 0 won't work

[thin venture]

so k doesnt matter

[swift orbit]

So let me get this straight.

[old nimbus]

bacc

[thin venture]

Other way

x^2 + R = 0 has real roots if R <= 0

then you need one constant

[thin venture]

but if R > 0 you get complex roots

[swift orbit]

Right. Ok. Let's assume that I have (x²+a)²

[thin venture]

and you need to do Cx+D

[swift orbit]

This person is now confusing me

[meager orbit]

Mb

[swift orbit]

I want to get my theory right

Not this question only

[thin venture]

He just showed you that for positive R there are no real solutions

but complex

[meager orbit]

Just the graphical method 😭

[thin venture]

it's good

the thing is

x^2+R is not reducible if R > 0 so we need now a linear term in the numerator if we apply partial fraction decomposition

[swift orbit]

Ooohh... For positive R.... Kkk

Thank you

I understand

Just. Clear this up for me k?

[thin venture]

clear this up?

[swift orbit]

Is this correct?

[swift orbit]

The main doubts

[thin venture]

how is a defined

[swift orbit]

A constant

[thin venture]

wow

[swift orbit]

Real constant

[thin venture]

is it real?

ok

then you need to do two cases

if a >= 0 you don't need the linear term

else you do

[swift orbit]

Riiiight

Assuming a is a negative, we won't get any real roots. So we simply keep the numerator a constant

[thin venture]

not necessarily integer

it can be a real positive number or 0

then you can reduce it x^2-a^2 = (x-a)(x+a)

[swift orbit]

Am I correct?

[swift orbit]

Is this correct?

@thin venture

[thin venture]

no

if you have complex roots you need a linear term in the numerator

not constant

[swift orbit]

And if I don't?

Constant

[thin venture]

ye

[swift orbit]

Ok. I'm done

Thank you sir bacc

How do I close this thread?

-solved

--solved

[thin venture]

!done

[summer bough]

If you are done with this channel, please mark your problem as solved by typing .close

[swift orbit]

.close

[summer bough]

Solved

Post marked as solved by @swift orbit.

Use .unsolved if this was a mistake.

[swift orbit]

I now understand

If a >=0 then the factors can be further broken into (x²-a) = (x-√a)(x+√a)

And each can have a numerator as a constant