Original message
| Time Mage
154th Post
 
Regular Customer
   
|  "Regarding the math problem"
, posted Mon 11 Feb 04:52
This is posted in the old BBS, with the original problem, but I want to post it here, so I'm sure it's viewed (I hate not correcting my mistakes).
OK, I've found the error. It's in equation (7):
7*y^2 + (17/24)*V*y - V^2/12 = 0 (7*)
When it should read:
7*y^2 + (17/12)*V*y - V^2/12 = 0 (7)
The denominator in the term that multiplies y is 12, not 24. This has happened to me because of not writing down what I think. Remember, mental calculations are dangerous! (And this kind of errors have cost me some exams, no joke... ).
Now, here goes the answer corrected:
Solving (7) (I used Mathematica, a good math program, because it's quicker, but it's easy to solve manually, anyway):
y = -V/4
and
y = V/21
The first answer hasn't physical meaning, because is a negative speed, and we are FILLING the pool, so the speed has to be positive. We pick answer 2, then.
From (4), we have the time that takes y to fill the pool:
t = V/y = V/(V/21) = 21 hours
the smallest pipe (x) fills the pool in:
t + 7 = 28 hours.
NOW it's correct. Thanks for correcting me.
Ah, and I also thik that this problem is incredible for a 12 years old kid. I did problems like this with 14 or 15 years, not 12, and they were difficult. (but not now)
"Layer upon layer make your mark now...Haste!"
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Replies:
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| Time Mage
164th Post
Regular Customer
| "Complex numbers!" , posted Wed 13 Feb 08:50
quote:
" Seeing as you're the math expert around here, I was hopin you could help me out on something...
12 + 17x^2 + 6x^4 = 0
* The ^ means exponent
I've tried to factor this several ways but I can't seem to get it...think you can figure it out? Your help would be much appreciated man
"
No wonder if you haven't been able to factor it, because, as Holiday has said, anything with a even numbered exponent will always be a positive number, so the answer is complex. And when I say complex, I mean a complex number, i.e., (real part + imaginary part), in the form (a + b*i), where i is the square root of -1. I hope you know this, if not it's confusing...
Well, the answer. For this calculations I don't use my brain, instead I use the computer's brain  . Using Mathematica, an awesome math program, the answers (the four anwers) are:
x1 = -i*sqrt(3/2)
x2 = i*sqrt(3/2)
x3 = -2*i/sqrt(3)
x4 = 2*i/sqrt(3)
Where sqrt(z) is the square root of z. Note that in the answers are given in the (a + b*i) form, and in this case, a = 0 for all of them. These are called "pure imaginary numbers", because they don't have real part.
Ah! And cool tag, btw!
"Layer upon layer make your mark now...Haste!"
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| Freeter 1767th Post
Silver Carpet V.I.P- Platinum Executive
  
    
    
    
| "complexing..." , posted Wed 13 Feb 09:15
quote:
" No wonder if you haven't been able to factor it, because, as Holiday has said, anything with a even numbered exponent will always be a positive number, so the answer is complex. And when I say complex, I mean a complex number, i.e., (real part + imaginary part), in the form (a + b*i), where i is the square root of -1. I hope you know this, if not it's confusing...
Yeah, I know it. I remember some of it back from high school.
quote:
"Well, the answer. For this calculations I don't use my brain, instead I use the computer's brain . Using Mathematica, an awesome math program, the answers (the four anwers) are:
x1 = -i*sqrt(3/2)
x2 = i*sqrt(3/2)
x3 = -2*i/sqrt(3)
x4 = 2*i/sqrt(3)
Where sqrt(z) is the square root of z. Note that in the answers are given in the (a + b*i) form, and in this case, a = 0 for all of them. These are called "pure imaginary numbers", because they don't have real part."
Well, unfortunately I have to use my brain, since I have to show how I arrive at those answers. Thanks for the solutions though.
quote:
"Ah! And cool tag, btw!
"
Thanks
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| Time Mage
167th Post
Regular Customer
| "The expanded solution." , posted Wed 13 Feb 11:02
quote:
" quote:" No wonder if you haven't been able to factor it, because, as Holiday has said, anything with a even numbered exponent will always be a positive number, so the answer is complex. And when I say complex, I mean a complex number, i.e., (real part + imaginary part), in the form (a + b*i), where i is the square root of -1. I hope you know this, if not it's confusing...
Yeah, I know it. I remember some of it back from high school.
quote:"Well, the answer. For this calculations I don't use my brain, instead I use the computer's brain . Using Mathematica, an awesome math program, the answers (the four anwers) are:
x1 = -i*sqrt(3/2)
x2 = i*sqrt(3/2)
x3 = -2*i/sqrt(3)
x4 = 2*i/sqrt(3)
Where sqrt(z) is the square root of z. Note that in the answers are given in the (a + b*i) form, and in this case, a = 0 for all of them. These are called "pure imaginary numbers", because they don't have real part."
Well, unfortunately I have to use my brain, since I have to show how I arrive at those answers. Thanks for the solutions though.
quote:"Ah! And cool tag, btw!
"
Thanks
"
Well, here goes the "expanded" solution:
We have :
12 + 17*x^2 + 6*x^4 = 0
If we make the change z = x^2, we have a cuadratic(spelling?) ecuation, and we know how to solve this:
12 + 17*x^2 + 6*x^4 = 0;
z = x^2;
12 + 17*z + 6*z^2 = 0;
We have transformed the equation in one that we can solve:
z = [-17 +-sqrt(17^2 - 4*6*12)]/(2*6) =
= (-17 +- sqrt(1))/12;
z1 = -18/12 = -3/2
z2 = -16/12 = -4/3
Now, we have x by:
z = x^2;
x = +-sqrt(z)
So, we have:
x1 = sqrt(-3/2) = i*sqrt(3/2)
x2 = -sqrt(-3/2) = -i*sqrt(3/2)
x3 = sqrt(-4/3) = i*2/sqrt(3)
x4 = -sqrt(-4/3) = -i*2/sqrt(3)
OK, now this answer should satisfy your needs, OK? Before I thought that you only needed the answers, sorry.
"Layer upon layer make your mark now...Haste!"
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"Re(1):HEY TIME MAGE! HELP!"