Sign in to follow this  

Math help.

This topic is 4844 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

If you intended to correct an error in the post then please contact us.

Recommended Posts

God this shit again. Simplify the expression :-/ x(x^2 + 5)^-5 + x^2(x^2+5)^-4 God this is hard can someone help direct me somehow? I know i have asked questions on this before but I really cant figure this one out. Plz help.

Share this post


Link to post
Share on other sites
ok the common things are (x^2+5).
EDIT:
I would think you would do something like this:
x(x^2+5)^-5 + x^2(x^2+5)^-4

((x^2+5)^-4) / ((x^2+5)^-5) = x^2+5
Then I think i would divide x^2 by x and that would be x so would it be:
x(x^2+5) = x^3+5x?

Share this post


Link to post
Share on other sites
What math course are you in..? I'd just raise the terms in the parentheses to the power, distribute in the x, and then add/subtract the terms. Your teacher should really be able to help you with this or a kid in your class.

Share this post


Link to post
Share on other sites
I'm in a precal class this is a prerequisite but I was never taught it and i'm trying to catch up. The teacher is lazy and never helps says she is busy.
x(x^2 + 5)^-5 + x^2(x^2+5)^-4
then we see that the like term is (x^2 + 5)^-5 then we divide it from both sides giving us:
x(1) + x^2 (x^2+5)
is that part right?
EDIT:Is the answer:
x^4 + 5x^2 + x?

Share this post


Link to post
Share on other sites
No. You can only divide through when equating it to something. What you do there doesn't work.

x(x^2 + 5)^-5 + x^2(x^2+5)^-4 != x(1) + x^2(x^2+5)
but
x(x^2 + 5)^-5 + x^2(x^2+5)^-4 = ((x^2 + 5)^-5)(x + x^2(x^2 + 5))

Are you sure you're working with the correct expression? It's awfully complex. The axiom program tells me the answer is

(5) -> x*((x^2 + 5)^(-5)) + (x^2)*((x^2+5)^(-4))

4 2
x + 5x + x
(5) -------------------------------------------
10 8 6 4 2
x + 25x + 250x + 1250x + 3125x + 3125


But that just looks like all it's done is expand it.

Share this post


Link to post
Share on other sites
Unfortuanately, since this doesn't equal anything you can't just divide off the like terms. You can only multiply or divide by one (or an expression that equals one), you can also add or subtract 0 (or expressions equaling zero). Because of this there aren't many options.

Try writing both values in rational form (as fractions), and then find the common demonimator.

Then rewrite as a single rational. Look for terms that can be reduced or rewritten into a form that allows for reduction (I didn't find any). The answer isn't pretty, but it is simplified.

Most of this is just mechanics, so make sure you understand the procedure of simplifying:

Write everything in standard notation (x^-n = 1/x^n)
Look for common terms
Rewrite in a more compact form
Repeat looking for common terms and simplfying and reducing
(Most importantly)--DO ONE STEP AT A TIME!!

As you gain proficiency you'll be able to group multiple simplifications together, but for practice or complicated problems, take it easy. Calculus is graded just as much on proper procedure as it is the correct answer.

Hope this helps...

Share this post


Link to post
Share on other sites
Start manipulating, using the rules of algebra.
When you begin to understand how they affect the expression, start moving the equation into the required form.

Share this post


Link to post
Share on other sites
Quote:
Original post by Vlion
Start manipulating, using the rules of algebra.
When you begin to understand how they affect the expression, start moving the equation into the required form.


Exactly...

Share this post


Link to post
Share on other sites
No. You can only divide through when equating it to something. What you do there doesn't work.
[quote]
x(x^2 + 5)^-5 + x^2(x^2+5)^-4 != x(1) + x^2(x^2+5)
but
x(x^2 + 5)^-5 + x^2(x^2+5)^-4 = ((x^2 + 5)^-5)(x + x^2(x^2 + 5))

Are you sure you're working with the correct expression? It's awfully complex. The axiom program tells me the answer is


(5) -> x*((x^2 + 5)^(-5)) + (x^2)*((x^2+5)^(-4))

4 2
x + 5x + x
(5) -------------------------------------------
10 8 6 4 2
x + 25x + 250x + 1250x + 3125x + 3125



But that just looks like all it's done is expand it.
[/QUOTE]
I don't think it's supposed to be this complex. I'm almost positive none of the problems we did in class looked like that. Although we only did 2 or 3 they were MUCH less complex.

Share this post


Link to post
Share on other sites
It just occured to me that I gave away the (probable) answer. instead I'll just show the first two steps to give him an idea of where to start.

First I rewrote it like this:

1 1
--------- + ----------
2 5 2 2 4
x(x + 5) x (x + 5)

Then I 'cross multiplied'. That is, with an equation like a/b + c/d, that equals (ad + bc) / bd, to get:

2 2 4 2 5
x (x + 5) + x(x + 5)
--------------------
3 2 9
x (x + 5)


I think if you get here you should be able to get the rest

Share this post


Link to post
Share on other sites
You forgot that the exponents were negatives.

Edit: Nevermind, I see what's happening. But, how did you get the x2 on the bottom? It's not being affected by the ^-4 so how does that work. This problem is a major waste of time. She could have taught the same material without giving such a drawn out problem.

Share this post


Link to post
Share on other sites
This is slightly off topic, but you should learn this:

to write x2 (or any exponents for that matter) you should use the <SUP> tags. It makes everything much nicer to read and easier to type!

Yay HTML!

Share this post


Link to post
Share on other sites
I tired simplfying it out and I got an answer which isnt like other stuff posted here so I dunno if its right. Anyway this is the sort of stuff we did in maths last year and its been while since I did it.

To start I suggest taking out common terms so from

x(x2 + 5)-5 + x2(x2 + 5)-4

the (x2 + 5)-5 and x are common to both (as they both have the lowest power). Put these out of brackets and whatevers left of the expression inside.

Im not sure how much more help you want (if ive helped you at all lol) but if you want more from me just ask :)

Share this post


Link to post
Share on other sites
Quote:
Original post by nobodynews
It just occured to me that I gave away the (probable) answer. instead I'll just show the first two steps to give him an idea of where to start.

First I rewrote it like this:

1 1
--------- + ----------
2 5 2 2 4
x(x + 5) x (x + 5)

Then I 'cross multiplied'. That is, with an equation like a/b + c/d, that equals (ad + bc) / bd, to get:

2 2 4 2 5
x (x + 5) + x(x + 5)
--------------------
3 2 9
x (x + 5)


I think if you get here you should be able to get the rest

actually the x and the x2 stay in the numerator...

...this leaves a demonimator of (x2+5), except to different powers, finding the common demonimator should be fairly easy.

Share this post


Link to post
Share on other sites







Hey ssjcory, I’ve put the complete solution but I’ve left big gaps between each step so that you can choose how much of it you want to read.

Firstly lets be real sure you know what’s meant by simplifying (coz the stuff you were suggesting suggested your understanding is a bit hazy ;) Basically you start with a hard looking expression and then you just try to make it look simpler keeping the expression identical at all steps. Here’s an easy simplifying example:

Simplify:

x + x + 5x + 2x + x4 / x2

Answer:

x + x + 5x + 2x + x4 / x2

= x + x + 7x + x4 / x2

= 2x + 7x + x4 / x2

= 9x + x4 / x2

= 9x + x2

I’ve deliberately made it real (almost stupidly) easy, just to remind the concept here (coz when the questions get harder you kind of forget or get confused of the underlying methodology and principles. Ok, now onto your problem:

Firstly do you know that x -5 is the same as 1 / x 5 ?



Ok, then you’ll realise that your expression is the same as:

x / (x 2 + 5) 5 + x 2 / ( x 2 + 5 ) 4



So that’s the first step of the simplification. The next step is to realise that the above is the same as this:


x / (x 2 + 5) 5 + x2 (x2 + 5) / (x2 + 5) 5



Now realise that both terms in the sum have the same denominator. So you can write it like this:


x + x2 (x2 + 5)
____________

( x2 + 5 ) 5

Share this post


Link to post
Share on other sites
I'm closing the thread immediately. How did I miss this one? I don't think it showed up in the list on my screen..

Folks, homework/schoolwork related questions are OFF TOPIC, with few exceptions.

Please go read or re-read the Forum FAQ.

There are websites DEDICATED to helping with math homework/schoolwork. Please in the future use THOSE sites for this type of question!

Share this post


Link to post
Share on other sites

This topic is 4844 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

If you intended to correct an error in the post then please contact us.

Guest
This topic is now closed to further replies.
Sign in to follow this