Hard Light Productions Forums
Off-Topic Discussion => General Discussion => Topic started by: Locutus of Borg on February 01, 2011, 03:49:19 pm
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Why
(20 - 14 * 2^(1/2))^(1/3) = 2 - (2)^1/2
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Erm, (P)lease (E)xcuse (M)y (D)ear (A)unt (S)ally?
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(20-14*root2)^1/3 = 2 - root2
root2 = 1.4142
(20 - 14 * 1.4142) ^ 1/3 = 2 - 1.4142
(20 - 19.7988) ^ 1/3 = .5858
(.2012) ^ 1/3 = .5858
.59 = .5858, the error's just introduced by rounding earlier
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How do I arrive at 2 - (2)^1/2 though
I need to write a proof starting with the first term.
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Battuta, I think any idiot can do that substitution and doesn't help OP to learn about symbolic calculations.
The easiest way to derive that expression is simply to cube both sides of the equation.
20-14*sqrt(2)=(2-sqrt(2))3
Now open the right hand side and remember that (a-b)3 = a3 - 3a2b + 3ab2 - b3
You can then collect the terms. More than that I'm not gonna say, and I may have already said too much.
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Battuta, I think any idiot can do that substitution and doesn't help OP to learn about symbolic calculations.
I'm well aware of that, but the OP didn't make it clear that he wanted anything more than a simple arithmetic resolution, nor did it suggest anything about learning about symbolic calculations. As far as we knew this was an order of operations confusion.
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Or, if you didn't know the polynomial expansion of (a - b)3, you might need to derive it there. But that should be easy.
ADDENDUM:
Let's try to put it down one more time: either derive the expansion of (a-b)3 , or substitute the numbers to (a-b)3 and simplify. I recommend deriving the expression, though, since if you get a job like mine, more often than not it is the derivations of the equations that are more important than just plunging in the numbers.
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My problem is that I figured out that the expression equals 2 - (2)^1/2 on a calculator. I need to explain how I got there from (20 - 14 * 2^(1/2))^(1/3)
This is all part of a larger question but once this is solved then the rest is simple.
(and yes I know that (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3
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I'd suggest following Mika's method. Don't plug anything into the calculator, nor try to evaluate the sqrt(2) until the very end.
You should essentially come out with an equation of sums on one or both sides of the equal sign.
Btw, I ended up with 20 = 20 :p
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Here's the entire question
if x = (20 - 14 * 2^(1/2))^(1/3) + (20 + 14 * 2^(1/2))^(1/3)
What is x^3
I know the answer is 64 but I can't PROVE it.
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I cubed both sides (of the original question) and expanded the right side. It seems kinda random, but I just figured exponents would be easier to work with than roots.
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But then you're still left with fractional exponents.
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Fractional exponents are fine, you can always split them into a power and a root if it's easier to look at for you.
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I figured out that the expression equals 2 - (2)^1/2 on a calculator. I need to explain how I got there
Thats easy just be truthful write as your answer "I typed it into a calculator"
Davros leaves the thread the audience are stunned into silence by his totally awesome mathematics prowess
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Disclaimer: I don't do home assignments for people, but I might help in solving them. Work needs to be done by the OP himself.
That being said, due to the perceived stupidity/geniosity (that's right) of this assignment, I might help you to do this, but first of all you need to show me you have really done something yourself.
(1) Post where you are at the moment, and what you have tried and why.
(2) Then tell me your age and what you know about counting with abstract letters instead of numbers.
I'm not trying to be rude, (1) is needed to convince me that you have indeed done something yourself. (2) is needed because I suspect this assignment is beyond the capabilities of the average person of your age.
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We did this kind of stuff as college freshmen (or heck high school junior/seniors) all the time. Maybe even earlier.
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I said that the 20 + 14sqroot(2) = a
and 20 - 14sqroot(2) = b
a^3 + 3a^2b + 3ab^2 + b^3
a^3 + b^3 = 40
3a^2b = 160 + 112sqroot(2)
that's where I am right now. I'm not working on it atm.
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A couple of additional questions, explain me why would you want to cube immediately the equation
x = (20-14*sqrt(2))^(1/3)+(20+14*sqrt(2))^(1/3)?
Also, please answer (2).
I know this is kind of grilling, but I need to make sure you have at least done something yourself.
We did this kind of stuff as college freshmen (or heck high school junior/seniors) all the time. Maybe even earlier.
If you were really allowed to use numbers in calculations like this during high school, I'm indeed pretty much speechless.
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I'm cubing it because I need find x^3 - 6
2: I know a little calculus, so I am old enough to be able to do this ;7
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The question is, how well can you calculate with polynomials that consist only of letters? That is, If you want more formal resolution of the problem, you need to be able to solve a set of two equations that only consist of letters. I initially thought you were around 14, would that be anywhere near correct?
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We did this kind of stuff as college freshmen (or heck high school junior/seniors) all the time. Maybe even earlier.
If you were really allowed to use numbers in calculations like this during high school, I'm indeed pretty much speechless.
Of course we did pure algebraic manipulations as well, both in pure math and in topics like special relativity. But numerical calculations come up all the time in physics and other mathematical applications too. It's not a question of whether or not you're allowed to do them, sometimes you need to.
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lol no
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I don't know to which question you answered, but here goes.
Why would you want to cube the nasty looking cubic root side? That's tedious work, I think your original approach was better. In fact, we would like to show that under that cubic root, there indeed is a cube of something. Why? Because that makes it rather easy to calculate and the cubic roots are cancelled.
So in principle, the effect should be [(a - b)3]1/3 = a - b
How could you know that 20-14*sqrt(2) is a cube of something? Well, you are actually cheated by the looks of it, since using numbers allows combining terms in a way that it is very non-intuitive to get back in to the expanded cubic form of a3 - 3a2b + 3ab2 - b3. Which is the reason I still heartily recommend staying the hell away from the numbers until last possible moment and said this is a rather arbitrary and stupid kind of assignment.
So now we are basically saying that a3 - 3a2b + 3ab2 - b3 = 20 - 14*sqrt(2)
That's fine, but not very informative. We have two unknowns a and b and only one equation. To solve the equation exactly, we would need two independent equations. Where do we get the second one?
Well, there is a bit of a hint in the numerical form, so we can actually write:
(1) a3 + 3ab2 = 20
(2) -3a2b - b3 = -14*sqrt(2)
Now, as I said I wont solve the problem myself, so spoon-feeding the answer stops here. Explain me why did I write these two equations; I wont continue any further until I get reasonable feedback from you. I'm positive that you will get it, but it might take some time. This is where the assignment actually becomes ingenious.
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Well since b is the radical then the terms that have it to an even power will result in an integer
the terms that have the radical to an odd power will have it as a radical still
Those two equations make a system. In order to solve it we need to make them both equal the same thing. (Adding or subtracting the equations gets us back where we started :@)
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Your deduction is basically correct.
There are a couple of manipulation steps for the two equations.
Let's mark b=C*sqrt(2), where C is a constant. Why? Because we know that the sqrt(2) has to be a participant of the equation (2), but we don't know for sure if it had a multiplier in front of it.
The equations turn to:
(1) a3 + 6aC2 = 20
(2) -3a2C - 2C3 = -14
And to make it a little bit more clean, multiply (2) with -1.
(1) a3 + 6aC2 = 20
(2) 3a2C + 2C3 = 14
Here, there are two options for continuation, either you might guess the solution for both equations, or, there is the analytical solution for them, but from this point on it gets rather complex.
I will not do further steps today as I need to go to sleep, if you really want to see how complex it can get, solve a as C's function. Let's see what's happened here tomorrow then.
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Turns out I copied the question wrong. It really asks for x^3 - 6x
if the contents under the first radical (the plus one) is a and the contents under the second radical (the minus one) is b then
x = a^1/3 + b^1/3
x^3 = a + a^2/3 b^1/3 + a^1/3 b^2/3 + b
a+b = 40
6x = 6a^1/3 - 6b ^ 1/3
x^3 - 6x = 40 + 3a^2/3 b^1/3 + 3a^1/3b^2/3 - 6a^1/3 - 6b^1/3
factor out (3)(A^1/3 + b^1/3) from the right
x^3 - 6x = 40 +3 (a^1/3 + b^1/3)(a^1/3 b^1/3 - 2)
a^1/3 + b^1/3 = x so
x^3 - 6x = 40 + 3x ((ab^1/3 - 2)
ab^1/3 = (400 - 391)^1/3 = 2
2 - 2 = 0!
x^3 - 6x = 40 + 0
x^3 - 6x = 40
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Yeah, I was kinda thinking that there has to be something wrong with the assignment, as solving the earlier one by the book would have required the formula for solving a third order polynomial... Perhaps even complex numbers.
Next time be more careful.
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Thanks for your help anyway :)