ahhh...

brings back memories, of walking out on my maths class (teacher didn't give a shit)...

some how i Still graduated with a B+....

i think there's a special way to get derivatives of fractions, or maybe fractions of powers.

if i find my old texts, i'll PM u....

but that probably won't happen.

we got the answer on page two guys, chill!!

ill probably come back to this thread with more problems, when i continue working later.

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quote:

Originally posted by Krypton we got the answer on page two guys, chill!! ill probably come back to this thread with more problems, when i continue working later.

another problem on the first page of the thread, post 1...

i know the answer to this one, but can u find it?

Find all the real RATIONAL zero's in this equation. (In other words, find how many time "x" crosses the "x" axis on a graph.)(or find all the real solutions of "x" that makes the equation = to zero. youll get a PRIZE!!!

x^3 - 4x^2 + x + 6 = 0

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Damn it, a third power!!! There's a method to do this but I forgot how, lemme ask my sis, she just passed algebra so it's still fresh in her head...

quote:

Originally posted by ::TranceVanDyk:: another problem on the first page of the thread, post 1... i know the answer to this one, but can u find it? Find all the real RATIONAL zero's in this equation. (In other words, find how many time "x" crosses the "x" axis on a graph.)(or find all the real solutions of "x" that makes the equation = to zero. youll get a PRIZE!!! x^3 - 4x^2 + x + 6 = 0

x =-1; x=2; x=3

or

(x+1)(x-2)(x-3)=0

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writing lyrics aint what it used to be

quote:

Originally posted by Jocker x =-1; x=2; x=3 or (x+1)(x-2)(x-3)=0

I'm at a loss...could you please show me how to do this or at least what method you used?

quote:

Originally posted by stevieboy32808 I'm at a loss...could you please show me how to do this or at least what method you used?

finding the first root (-1) with picking of numbers (you go 0, +-1, +-2, +-3, etc... 99.9% of cubic equations in college textbooks have at least one real root in the +-3 range). then, factoring it out, and solving the quadratic equation. there is a much more complex formula for solving cubic equations (at least for real numbers, i'm not sure about the complex ones), but i don't remember it

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writing lyrics aint what it used to be

So why the difference of answers...

here's one:
disprove the pythagorean theorem.

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quote:

Originally posted by Jocker x =-1; x=2; x=3 or (x+1)(x-2)(x-3)=0

niceee

quote:

I'm at a loss...could you please show me how to do this or at least what method you used?

x^3 - 4x^2 + x + 6

Possible zeros: +-6, +-3, +-2, +-1 (all factors of c/all factors of the 1st coefficent

guess and check. find out that when u divide 3 by x^3 - 4x^2 + x + 6 there is no remainder. so u know x-3 is a factor and that 3 is one solution. after dividing by 3, u get x^2 - x - 2. factor that out, and u get (x-2)(x+2) and then u include the (x-3).

use the zero theorem or whatever its called
x-2 = 0 (x = 2)
x+1 = 0 (x = -1)
x-3 = 0 (x = 3)

u have your answers.

to get the complex solutions if there are any, i believe u use the quadratic equation once u find at least one real solution. i have no idea how to find solutions to an equation with all complex zeros.
x-3 = 0 (

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you have a fundamental flaw

x(x^2-4x+1)=-6 doesn't mean that x=-6 and x^2-4x+1=-6 (judge yourself, then the answer would be -6*-6); you can say that different factors on the left equal the right side only when the right side is equal to 0. i.e x(x^2-4x+1)=0 would indeed mean that x=0 and x^2-4x+1=0

edit: this is to stevieboy

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writing lyrics aint what it used to be

quote:

Originally posted by kadomony here's one: disprove the pythagorean theorem.

only if not in euclidian geometry, because in euclidian geometry (i.e. geometry that you study in school) the pythagorean theorem holds true.

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writing lyrics aint what it used to be