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The Math Thread (4 N3rdS!!!) (pg. 4)
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| Abhay |
ahhh...
brings back memories, of walking out on my maths class (teacher didn't give a )...
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.:clown: |
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| ::TranceVanDyk:: |
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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| ::TranceVanDyk:: |
| 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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| stevieboy32808 |
| 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... |
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| Jocker |
| 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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| stevieboy32808 |
| 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? |
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| Jocker |
| 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:p |
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| stevieboy32808 |
So why the difference of answers...
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| kadomony |
:wtf:
here's one:
disprove the pythagorean theorem. |
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| ::TranceVanDyk:: |
| 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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| Jocker |
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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| Jocker |
| quote: | Originally posted by kadomony
:wtf:
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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