# Math :D

Registered User regular
edited May 2009
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.

MKR on

## Posts

• Registered User new member
edited April 2009
Very first line, dividing 2x^4 by 2x^2 is just x^2, not 2x^2.

Sandrock on
• Registered User regular
edited April 2009
It's always the blindingly obvious things.

Thanks.

MKR on
• Registered User regular
edited April 2009
you should be able to multiply what's on top by what's to the left and get what's underneath.

step 1
__x^2_____________
2x^2+x+1
l 2x^4 -3x^3 +x +1
-(2x^4 +x^3 +x^2)
-4x^3+x^2+x+1
step 2
__x^2_-2x__________
2x^2+x+1
l 2x^4 -3x^3 +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

Dman on
• Registered User regular
edited April 2009
Yeah, I just forgot that n/n=1. The problem is nice and solved over on the next page, and I should ace the test I'm doing this afternoon.

MKR on
• Registered User regular
edited April 2009
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 gets around to answering my questions by e-mail, but I don't think they pay him enough to be full time, so I tend to 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 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.

MKR on
• Registered User regular
edited April 2009
MKR wrote: »
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 gets around to answering my questions by e-mail, but I don't think they pay him enough to be full time, so I tend to 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 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.

2 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-99 on
• Registered User regular
edited April 2009
DJ-99 wrote: »
2 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.

I 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.

MKR on
• San DiegoRegistered User regular
edited April 2009
Add the exponents, the same as if they weren't fractions.

So
x^(1/3) * x =
x^(1/3) * x(3/3) =
x^(4/3)

Orogogus on
• Registered User regular
edited April 2009
Orogogus wrote: »
Add the exponents, the same as if they weren't fractions.

So
x^(1/3) * x =
x^(1/3) * x(3/3) =
x^(4/3)

It 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.

MKR on
• San DiegoRegistered User regular
edited April 2009
MKR wrote: »
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.

8x^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.
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.

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.

Orogogus on
• Registered User regular
edited April 2009
I should probably be able to see it, but * where is the 2x^1/3 pulled from?

edit: 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?

MKR on
• San DiegoRegistered User regular
edited April 2009
The 2 is probably self explanatory.

The 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.

Orogogus on
• Registered User regular
edited April 2009
I caught it just as you posted.

I'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.

MKR on
• Registered User regular
edited April 2009
Ok, one more fun thing on that problem.

4x^1/3

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

MKR on
• San DiegoRegistered User regular
edited April 2009
I'm not sure why your x^1/3 is becoming x^0/3 and x^0/3. As noted earlier, dividing is a matter of subtracting exponents. So you're pulling out the 2 and the entire x^1/3, and leaving 2 and x^0/3 (i.e., 1).

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

Orogogus on
• Registered User regular
edited April 2009
So it was blindingly obvious like I suspected. x^1/3 - x^1/3 = nada, so 4x^1/3 divided by the common factor of 2x^1/3 is 2.

Now I feel silly, and educated.

MKR on
• Registered User regular
edited April 2009
One more thing on that:
(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.

MKR on
• Registered User regular
edited April 2009
x^a * x^b = x^(a + b)

In 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.

Zek on
• Registered User regular
edited April 2009
So when it's a polynomial inside the parenthesis, I can't treat it like (a^n)^m = a^nm and use the power outside the parenthesis on each monomial inside?

MKR on
• Registered User regular
edited April 2009
Nope. If you want to expand it you have to do it the old fashioned way.

Zek on
• Registered User regular
edited April 2009
Old fashioned way? You mean factoring?

And 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.

MKR on
• Registered User regular
edited April 2009
I mean you just need to separate it out and do the multiplication, i.e. FOIL for binomials.

(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.

Zek on
• Registered User regular
edited April 2009
The two weeks of this class have been a bit of a roller-coaster, so that's probably why I'm not spotting stuff like that which should be obvious. My brain is still sizzling.

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

Thanks for the help.

MKR on
• Registered User regular
edited April 2009
it really helps me to the put the square root symbol over the 1/2s and whatnot. A lot easier to see the algebra.

Demerdar on
• Registered User regular
edited May 2009
New problem:

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.

MKR on
• Registered User regular
edited May 2009
MKR wrote: »
New problem:

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.

Part 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.

JHunz on
Gamertag: JHunz. R.I.P. Mygamercard.net
• Registered User regular
edited May 2009
JHunz wrote: »
MKR wrote: »
New problem:

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.

Part 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.

But why is that being done? And which of the things mentioned are "them?"

MKR on
• San DiegoRegistered User regular
edited May 2009
I think the reason they go out of their way to put x^2 on the bottom in the radical is in the text below the equations -- something happens as x approaches positive or negative infinity (y would get really big and really small, but I don't know if that's what they're getting at).

Them: Multiplying x^2/a^2 by a^2/x^2 gives you 1.

Orogogus on
• Registered User regular
edited May 2009
Orogogus wrote: »
I think the reason they go out of their way to put x^2 on the bottom in the radical is in the text below the equations -- something happens as x approaches positive or negative infinity (y would get really big and really small, but I don't know if that's what they're getting at).

Them: Multiplying x^2/a^2 by a^2/x^2 gives you 1.

It 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.

MKR on
• Registered User regular
edited May 2009
MKR wrote: »
Orogogus wrote: »
I think the reason they go out of their way to put x^2 on the bottom in the radical is in the text below the equations -- something happens as x approaches positive or negative infinity (y would get really big and really small, but I don't know if that's what they're getting at).

Them: Multiplying x^2/a^2 by a^2/x^2 gives you 1.

It 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.

The 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.

Dragoran on
• Registered User regular
edited May 2009
MKR wrote: »
Orogogus wrote: »
I think the reason they go out of their way to put x^2 on the bottom in the radical is in the text below the equations -- something happens as x approaches positive or negative infinity (y would get really big and really small, but I don't know if that's what they're getting at).

Them: Multiplying x^2/a^2 by a^2/x^2 gives you 1.

It 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.

So 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?

Demerdar on
• Registered User regular
edited May 2009
Demerdar wrote: »
MKR wrote: »
Orogogus wrote: »
I think the reason they go out of their way to put x^2 on the bottom in the radical is in the text below the equations -- something happens as x approaches positive or negative infinity (y would get really big and really small, but I don't know if that's what they're getting at).

Them: Multiplying x^2/a^2 by a^2/x^2 gives you 1.

It 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.

So 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?

Asymptotes 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.

Clipse on
• Registered User regular
edited May 2009
Demerdar wrote: »

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

I 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
The outside number has been multiplied by x^2/a^2, and the inside number has been multiplied by a^2/x^2.

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

edit: I guess I know how, but not why.

MKR on
• Starting to get dizzy Registered User regular
edited May 2009
The distributive property states that x(y + z) = (xy + zy). Usually you transform an expression from the left form of that to the right one, but you can also go the other way around. In particular, if you divide out the factor x from every term in the parenthesis, you can then multiply the whole thing by x without changing the overall expression's value.

For 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.

Smasher on
• Registered User regular
edited May 2009
Smasher wrote: »
The distributive property states that x(y + z) = (xy + zy). Usually you transform an expression from the left form of that to the right one, but you can also go the other way around. In particular, if you divide out the factor x from every term in the parenthesis, you can then multiply the whole thing by x without changing the overall expression's value.

For 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.

There 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.

MKR on