stumped on math problem?
whttrshpunk
03-12-2003, 02:12 PM
omg this has to be soooo simple but i cant figure it out. PLEASE HELP ME
x^2/3-1x^1/3-2=0
wth? this makes no sense to me....no comprende....someone help me please dear god
x^2/3-1x^1/3-2=0
wth? this makes no sense to me....no comprende....someone help me please dear god
Rod Himself
03-12-2003, 02:25 PM
put some parenthesis to group it up. its hard to tell without them in there.
could be
(x^2)/3-(1x^1)/3-2=0
or
(x^2/3)-(1x^1/3)-2=0
or some combination like that. its just hard to tell without them.
could be
(x^2)/3-(1x^1)/3-2=0
or
(x^2/3)-(1x^1/3)-2=0
or some combination like that. its just hard to tell without them.
whttrshpunk
03-12-2003, 02:26 PM
x^(2/3)-1x^(1/3)-2=0
Jay!
03-12-2003, 02:39 PM
I know the answer, but haven't found the right way to explain it yet...
why is it "1x" in the second part? Is that a typo, or irrelevant, or what?
why is it "1x" in the second part? Is that a typo, or irrelevant, or what?
whttrshpunk
03-12-2003, 02:40 PM
its irrelevant...sorry.....anyhow i dont know how to factor a quadratic equation with fractional exponents... :confused: :confused:
crab
03-12-2003, 02:45 PM
yes it is very simple
x^(2/3)-1x^(1/3)-2=0
x^3 * [x^(2/3)-1x^(1/3)-2] = 0 * x^3
x^2 - x - 2x^3 = 0
2x^3 - x^2 + x = 0
x * (2x^2 -x + 1) = 0
now we know x = 0 is a sol'n
As for 2x^2 - x + 1 = 0, use the quadratic formula to get
[1 +/- root(1 - 4 * 2 * 1)] / [2 * 2] these are imaginary roots
x^(2/3)-1x^(1/3)-2=0
x^3 * [x^(2/3)-1x^(1/3)-2] = 0 * x^3
x^2 - x - 2x^3 = 0
2x^3 - x^2 + x = 0
x * (2x^2 -x + 1) = 0
now we know x = 0 is a sol'n
As for 2x^2 - x + 1 = 0, use the quadratic formula to get
[1 +/- root(1 - 4 * 2 * 1)] / [2 * 2] these are imaginary roots
Jay!
03-12-2003, 02:48 PM
x^(2/3) = (cube root of x)squared
-1x^(1/3) = negative(one times (cube root of x))
-2 = negative two
i don't know where to get all the characters to type this out correctly... :o
anyways, treat "cube root of x" as "y", i.e. it's own independent variable. solve for "cube root of x", then you can get x
-1x^(1/3) = negative(one times (cube root of x))
-2 = negative two
i don't know where to get all the characters to type this out correctly... :o
anyways, treat "cube root of x" as "y", i.e. it's own independent variable. solve for "cube root of x", then you can get x
Jay!
03-12-2003, 02:50 PM
Originally posted by crab
now we know x = 0 is a sol'nIf x = 0, then x^(2/3)-1x^(1/3)-2 = -2
(right?)
now we know x = 0 is a sol'nIf x = 0, then x^(2/3)-1x^(1/3)-2 = -2
(right?)
whttrshpunk
03-12-2003, 02:54 PM
ok...heres the latest attempt at it
x^(2/3)-3x^(1/3)-10=0
the answers are 125 and -8...but i have no clue how they came up with that crap(i'm taking a math quiz online, the problem changes every time)
x^(2/3)-3x^(1/3)-10=0
the answers are 125 and -8...but i have no clue how they came up with that crap(i'm taking a math quiz online, the problem changes every time)
Toksin
03-12-2003, 02:58 PM
:cry::bloated: My head hurts:bloated::cry:
whttrshpunk
03-12-2003, 03:02 PM
w00000t! funny quiz =) The problems change every time you take the quiz(you can take it as many times as you want). well i lucked up and that problem came up twice, so i already had the answer written down. :D i=teh win 100! :cool:
Edit: Thanks for the help guys, I still want to figure out how to do it but now its time to go work out and release my frustration. :)
Edit: Thanks for the help guys, I still want to figure out how to do it but now its time to go work out and release my frustration. :)
Twist
03-12-2003, 07:30 PM
you taking the quiz on WebCT?
Oz
03-13-2003, 03:39 AM
Originally posted by Twist
you taking the quiz on WebCT?
Teh Oz hates teh WebCT quizzes. :(
you taking the quiz on WebCT?
Teh Oz hates teh WebCT quizzes. :(
whttrshpunk
03-13-2003, 10:22 AM
no its called mymathlab.....who cares i have straight 100s on all my quizzes so far...1 exam a 90 and 1 exam an 89...i sleep in that class every day :D
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