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MKR
Registered User regular

Since this book is riddled with non-example examples that don't really explain the problem, I'm going to make this a running math thread. My professor usually explains the problems well enough, but I don't think they pay him enough to be full time, so I generally go a full day without an answer.

PA has a lot of math-capable people on at a lot more times of the day, so it's the perfect place for me to get help.

Current problem(s):

I'm with it up to the third part. If I pretend that the fourth part doesn't exist, the stuff after that makes sense. Though I'm not sure why the radical in the last part doesn't cover the rational number, or why powers of 2 in it vanish.

It seems to be doing some wild algebraic voodoo where it flips the terms inside the parenthesis, inverts the rational number, and tosses some terms on to the rational number outside.

PA has a lot of math-capable people on at a lot more times of the day, so it's the perfect place for me to get help.

Current problem(s):

I'm with it up to the third part. If I pretend that the fourth part doesn't exist, the stuff after that makes sense. Though I'm not sure why the radical in the last part doesn't cover the rational number, or why powers of 2 in it vanish.

It seems to be doing some wild algebraic voodoo where it flips the terms inside the parenthesis, inverts the rational number, and tosses some terms on to the rational number outside.

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## Posts

SandrockonThanks.

MKRonstep 1

-(2x^4 +x^3 +x^2)

-4x^3+x^2+x+1

-(2x^4 +x^3 +x^2)

-4x^3+x^2+x+1

-(-4x^3 -2x^2 -2x)

I had this looking beautiful but my spaces don't show up so now it's fugly, hopefully you can still follow

DmanonMKRonPA has a lot of math-capable people on at a lot more times of the day, so it's the perfect place for me to get help.

Current problem(s):

I get all the way to the third from the last line, but I don't know how it goes to the next step. Getting from the second to last to the last is easy enough, but there's an intermediate step I'm missing. I also don't see why it's marking the twos out. It doesn't appear to be factoring them out since the 2 doesn't appear outside a parenthesis.

And I don't even know where to start on this one:

Where is it getting those common factors from? I know how to rewrite it as a problem with 3 as a denominator, but past that it's not clear.

MKRon2 is a common factor because it goes into both 4 and 6. x^(1/3) is a common factor because it goes (obviously) into x^(1/3) and if you multiply it by x, you get x^(4/3).

I'm sure you would see this if it said "Factor 4x^2+2x^4". Then the common factor would be 2x^2. You just have to get used to it when it's fractions, not whole numbers, in the exponent.

DJ-99onI think this is what's getting me. The book hasn't said anything about fractional exponents beyond converting to and from radicals. I don't know how to multiply x^(1/3) by x.

MKRonSo

x^(1/3) * x =

x^(1/3) * x(3/3) =

x^(4/3)

OrogogusonIt took me a moment to figure out where the 3/3 came from.

Now I get:

8x^4/3 + 4x + 6x^4/3

14x^4/3 + 4x

I'm having a brain fart and can't figure out how to factor it, but I'm sure that will come to me. Now, how about the first problem? It's likely that it's the same general principle and I just need to run over it again with what I know now, but I'll ask about it just in case it isn't a similar issue.

MKRon8x^4/3 + 4x + 6x^4/3

should be

8x^4/3 + 4x^1/3 + 6x^4/3

And then you can pull out the 2x^1/3 as described earlier.

Since you have two of (x^2 + 1) to the 1/2 being multiplied by each other you can do 1/2 + 1/2 on the exponents to get (x^2 + 1)^1, or just (x^2 + 1).

And the twos cancel out because you're basically multiplying 1/2 by 2, leaving you with 1/1, or 1.

Orogogusonedit: Wait, I think I see it.

x^4/3 is just x^1/3 * x^1/3 * x^1/3 *x^1/3, isn't it?

MKRonThe x^1/3 is pulled out of the x^4/3 and x^1/3.

You add exponents when multiplying, and subtract them to divide. So x^4/3 can be factored into x^1/3 and x^1.

Edit: Yes, that works, too.

OrogogusonI'm getting a better handle on this. I was worried I wouldn't be able to get all this before the test on Monday. PA people are the best people.

edit: Also, shouldn't the exponent on 4x be ^3/3?

edit: No, I just distributed incorrectly.

MKRon4x^1/3

Wouldn't this become 2x^0/3 * 2x^0/3? I'm probably looking at it wrong.

MKRonYou can try replacing x with something with a cube root like 8 or 27 to get a better feel for what's going on.

OrogogusonNow I feel silly, and educated.

MKRon(x^2 + 1)^1/2 (x^2 + 1)^1/2

You said to multiply these, but I'm a bit confused. If I do what I thought I was supposed to do with an exponent outside a parenthesis, I get (x + 1)(x + 1) (since ^2 * ^1/2 = 1), which would become something somewhat different from the answer if I use FOIL on it.

MKRonIn this case:

x = (x^2 + 1)

a = 1/2

b = 1/2

You can't modify the inside of the parentheses like that - you're ignoring the +1. Thinking about it another way, ^1/2 is simply a square root. Multiplying a square root with itself by definition results in the number contained within.

ZekonMKRonZekonAnd this book is as prone to omitting mentions of stuff like where certain things can't be used as it is to half-assing examples.

I was planning to write a basic math book after finishing the remedial class to get into this one, but now I think I'm going to expand the focus to college algebra once I finish this course and get good with using what I learned.

MKRon(x^2 + 1)^2 = (x^2 + 1)*(x^2 + 1) = x^2*x^2 + x^2 + x^2 + 1 = x^4 + 2x^2 + 1.

(x^2 + 1)^3 = (x^4 + 2x^2 + 1)*(x^2 + 1), etc...

In the case of square roots odds are there isn't a clean solution so you just need to leave the parentheses intact. In some cases it could be factored, like if you were taking the square root of the above result.

ZekonI think past this chapter it should be a little smoother.

Thanks for the help.

MKRonDemerdaronI'm with it up to the third part. If I pretend that the fourth part doesn't exist, the stuff after that makes sense. Though I'm not sure why the radical in the last part doesn't cover the rational number, or why powers of 2 in it vanish.

It seems to be doing some wild algebraic voodoo where it flips the terms inside the parenthesis, inverts the rational number, and tosses some terms on to the rational number outside.

MKRonPart 4: The outside number has been multiplied by x^2/a^2, and the inside number has been multiplied by a^2/x^2. Since multiplying them together gives you 1, you don't need to do anything to the other side of the equation.

Part 5: It's doing a square root of the entire side. The radical doesn't cover the left term because removing the powers of two has already taken care of the square root of that term.

JHunzonBut

whyis that being done? And which of the things mentioned are "them?"MKRonThem: Multiplying x^2/a^2 by a^2/x^2 gives you 1.

OrogogusonIt says something about using the asymptotes to help graph the equation, but I haven't done enough read-throughs of the chapter to make sense of it.

MKRonThe x goes on the bottom because of the way the function 1/(x^2) behaves--as x becomes large, 1/(x^2), and by extension a/(x^2), becomes infinitesimally small. It looks like it's looking at what happens when x starts approaching infinity, so as this happens, the a/(x^2) term becomes negligible, and the radical becomes effectively sqrt(1). Thus, you can effectively ignore it when looking at the equation.

What I'm about to say may be wrong since it's been a while, but I'm pretty sure this means that as x becomes large, the behavior of the hyperbola will start looking like the line y=(+ or -)bx/a. Thus, you can graph the lines y=bx/a and y=-bx/a (these lines are the asymptotes), and draw a hyperbola along them (you'll need to plug in some points to figure out where exactly it goes) to approximate the appearance of the function.

DragoranonSo an asymptote is really just where y->infinity as x approaches some number or vice versa

x approaches infinity.. y will approach some constant, and level off, thus creating an asymptote.

This really has nothing to do with what they did algebraically.

In step four, all they did was take out an x^2/a^2 from step 3. This is so that when they square rooted both sides, it would come out a bit nicer and more compact, and clearly show where the function is defined.

Maybe you can be a little more specific into what you aren't understanding in step 4?

DemerdaronAsymptotes don't have to be constants. For instance we can say the function f(x) = (x^2)/(2x + 1) is asymptotic to x/2 as x goes to infinity, and -x/2 as x goes to negative infinity.

ClipseonI don't know how it's going from b^2(x^2/a^2 - 1) to b^2x^2/a^2(1 - a^2/x^2).

JHunz said

But I don't know why that's happening.

edit: I guess I know

how, but notwhy.MKRonFor your example, we start with b^2 * [(x^2)/(a^2) - 1]. We divide each of the terms in the brackets by (x^2)/(a^2). Dividing (x^2)/(a^2) by itself is naturally 1. Dividing by a number is the same as multiplying by its inverse, so 1/[(x^2)/(a^2)] = 1*(a^2)/(x^2) [note that x and a switched]. We then multiply the parenthetical expression by (x^2)/(a^2) to counteract dividing the individual terms, and we get b^2 * (x^2)/(a^2) * (1 - (a^2)/(x^2)), which is what you find in the book.

SmasheronThere we go. And now I see why they did it that way. It produces a much simpler and easier equation to work with than if I had just distributed the original. This book never mentions why it does stuff like that. Fortunately, I haven't run into a confusing spot like that in over a month.

And if I square the resulting equation to reverse it, it makes the removal of powers in the term outside the parenthesis much clearer and more familiar.

MKRon