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General Discussion => Other Discussions => Math and Science => Topic started by: Yeong on March 10, 2011, 07:22:44 am

Title: New sense of quad formula! Factoring quad trinomial that haz imaginary root
Post by: Yeong on March 10, 2011, 07:22:44 am

I don't know if people already knows or just doesn't care, but imma post this anyway because I'm pretty sure this is help at least one people.  :)

Sum of two squares

This is not trinomials but anyway...
It's just similar to difference of two squares, but you just put i at the end

Ex) (x²-4) = (x-2)(x+2)
     (x²+4) = (x-2i)(x+2i)

trinomials

*only works when a=1
There is a step so...
Ex) x² -4x +13

1) (x     )(x     )
2) now put b/2.   (x-2    )(x-2    )
                                ________
3) now put plus/minus\/c-(b/2)²     x i   (x-2+3i)(x-2-3i)        

Title: Re: Factoring quad trinomial that haz imaginary root
Post by: phenomist on March 10, 2011, 07:09:43 pm

Yeah, that's the quadratic formula, with a=1.

The "formal" way would be to complete the square.

x^2 -4x + 4 + 9 = 0
(x-2)^2 = -9
x-2 = +3i or -3i
x = 2+3i or 2-3i
So those are the solutions, so necessarily (x-2-3i)(x+2+3i) = original trinomial, since same leading term.

Title: Re: Factoring quad trinomial that haz imaginary root
Post by: Yeong on March 10, 2011, 07:13:40 pm

I know, but I was just presenting other way.
Personally, I prefer quad formula better  ;D
Also, I dont think sqrt(c-(b/2)²) is not quad formula at a=1.
it should be -b(+/-)sqrt(b²-4c)

Title: Re: Factoring quad trinomial that haz imaginary root without using quad formula
Post by: phenomist on March 10, 2011, 07:18:36 pm

Well, you divide by 2 when a=1 (denominator's /2a for quad formula), so this gives you b/2, and a divide by 2 = a divide by 4 inside a sqrt sign, hence (b/2)^2-c. The i occurs because you extracted out one factor of sqrt(-1).

Title: Re: Factoring quad trinomial that haz imaginary root
Post by: Yeong on March 10, 2011, 07:33:54 pm

how does sqrt((b/c)^2-c) turns into (b^2 - 4c)?

Title: Re: Factoring quad trinomial that haz imaginary root
Post by: phenomist on March 10, 2011, 07:42:12 pm

original
[-b+/-rt(b^2-4ac)]/2a

Let a=1


[-b+/-rt(b^2-4c)]/2

-b/2 +/- rt(b^2-4c)/2

-b/2 +/- rt((b^2-4c)/4)

-b/2 +/- rt(b^2 / 4 - c)

- b/2 +/- rt((b/2)^2 - c)

- b/2 +/- i*rt(c-(b/2)^2)

Meh, I usually omit the sqrt( sign because it's annoying to type. It should be self-evident though.

Title: Re: Factoring quad trinomial that haz imaginary root
Post by: Yeong on March 10, 2011, 07:46:34 pm

ahh okee  :)
better change the title then.