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Author Topic: New sense of quad formula! Factoring quad trinomial that haz imaginary root  (Read 4650 times)

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Offline Yeong

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New sense of quad formula! Factoring quad trinomial that haz imaginary root
« 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)        

« Last Edit: March 10, 2011, 07:46:58 pm by yeongJIN_COOL »
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Offline phenomist

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Re: Factoring quad trinomial that haz imaginary root
« Reply #1 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.
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Offline Yeong

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Re: Factoring quad trinomial that haz imaginary root
« Reply #2 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)
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Offline phenomist

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Re: Factoring quad trinomial that haz imaginary root without using quad formula
« Reply #3 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).



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Offline Yeong

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Re: Factoring quad trinomial that haz imaginary root
« Reply #4 on: March 10, 2011, 07:33:54 pm »
how does sqrt((b/c)^2-c) turns into (b^2 - 4c)?
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Offline phenomist

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Re: Factoring quad trinomial that haz imaginary root
« Reply #5 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.
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Offline Yeong

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Re: Factoring quad trinomial that haz imaginary root
« Reply #6 on: March 10, 2011, 07:46:34 pm »
ahh okee  :)
better change the title then.
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