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Aug. 8th, 2006 @ 12:43 pm Hello & Quadratic Formula Expansion
Hey, BOGWarrior89 here, just like to say I'm out of high school and going into a college mathematics major program (alongside another major in physics). I took a semester of Pre-Calculus and Trigonometry, other than that, not much more to say on that subject.

My current "work" is on the quadratic formula (I had a problem with solving asymptotes using "x approaches infinity" limits). Unfortunately, I don't have my notebook with me, but I can try to recall it offhand. I seriously need an opinion on this, so here goes (new variables in bold):


" y = ax^2 + bx + c" -> " x = ( -b +- [ b^2 + 4ay - 4ac ]^1/2)/(2a) "

" y = ( ax^2 + bx + c )/( dx + e ) " -> " x = ( -b + dy +- [ b^2 + (d^2)(y^2) + 4aey - 2bdy - 4ac]^1/2)/(2a)"

" y = ( ax^2 + bx + c )/( fx^2 + dx + e ) " -> " x = ( -b + dy +- [ b^2 + (d^2)(y^2) + 4fey^2 + 4aey - 4cfy - 2bdy - 4ac]^1/2)/(2a - 2fy)"

The best part? If "y = 0", then the outcome for all three of these is the original quadratic formula.

I'm hoping I can post a picture of the real thing; it's easier to read. Tell me what you think.

BOGWarrior
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bogwarrior89:
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From:quatranoctal
Date:August 9th, 2006 12:50 pm (UTC)
(Permanent Link)
Interesting, but unfortunately unnecessary.

y = ax^2 + bx + c
ax^2 + bx + (c-y) = 0

Now you can, if you want, create new coefficients a' = a, b' = b, and c' = c-y, and then you just apply the good old quadratic formula to a'x^2 + b'x + c' = 0. Likewise for the others, only you have to move around the other coefficients too.

However, the general idea of using the quadratic formula when the coefficients aren't actually constant is a good thing to do - there are even applications on that kind of thing in more advanced ideas like asymptotic theory.

And, incidentally, you can apply it even if you're not working in x, but in something more complicated - eg.

e^(2x) - 4e^x + 3 = 0 can be written as
(e^x)^2 - 4e^x + 3 = 0 which gives
e^x = (4 +- (16 - 12)^1/2)/2
= 2 +- 1 = 1, 3
So x = 0, ln 3
From:bogwarrior89
Date:August 10th, 2006 06:01 am (UTC)
(Permanent Link)
Yes, indeed interesting; I already knew it would be "unnecessary"; just re-read the "best part" section. However, it is infact a little useful. My Pre-Calculus teacher used the x^2 polynomial/x^2 polynomial to solve a problem he had solved for us using limits earlier. The book said it was "4/3x" or something like that, and the above formula returned something like "4/3". Besides, it's good to get experience.