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.
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.
Math help.
Author
No. You can only divide through when equating it to something. What you do there doesn't work.
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:
I think if you get here you should be able to get the rest
First I rewrote it like this:
1 1--------- + ---------- 2 5 2 2 4x(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 5x (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
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.
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.
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!
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!
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 :)
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 :)
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 4x(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 5x (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.
Author
What is wrong with my method i'm almost positive that this was around the same way we did it and borodog how did u get thsoe numbers? [CONFUSSED]
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
This topic is closed to new replies.
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