# Anyone good with Algebra???

Discussion in 'Community Discussion' started by rdf8585, Jul 6, 2006.

1. ### rdf8585 macrumors regular

Joined:
Feb 15, 2006
#1
(3*sqrt(2) - 4)^2

(x^2 - x) / (x^2 - 1) divided by (x^2 - 2x) / ( x^2 + 3x + 2)

2x^-1y^2z^3)^-3(3x^2y^-3z^-4)^2 answer should contain only positive exponents

I really have no idea where to even start there. We had this big ole thing to do and these three are the only ones I'm clueless on. Seeing how I already failed the class once, I'm a bit desperate here. I look at these and go

So some help would REALLY be appreciated. Help a fellow Mac-head out.

2. ### jsw Moderator emeritus

Joined:
Mar 16, 2004
Location:
Andover, MA
#2
Am I missing something, or is this just (3*sqrt(2) - 4) * (3*sqrt(2) - 4)?

Surely you can expand (a - b) * (a - b) and then get the results.

The second one, if I read it correctly, is (a/b) / (c/d), which would be
(a/b) * (d/c), or ((a * d)/(b * c))... expand and see what you get.

And your third equation seems to have a typo or two in it... I can't seem to make sense out of it. Could you add parenthesis and spaces (and numbers) as appropriate?

3. ### ChrisBrightwell macrumors 68020

Joined:
Apr 5, 2004
Location:
Huntsville, AL
#3
For future reference: x^(-1) = 1 / x

I think this is the right approach.

if
a = 3 * sqrt(2) (since this is a single real value, approx 4.24)
b = 4​

then
( a - b )^2 = a^2 - 2ab + b^2​

4. ### ToddW macrumors 6502a

Joined:
Feb 26, 2004
#4
2
(x+2)/(x-2)
(y^2z^3)/1 - (27x^4)/3y3z^2

z^-x=1/z^x that is what it means by must have postive exponents

you should work those problems until you get those answers. that is if the answers are right. i just looked at them for a minute or so.

5. ### Timepass macrumors 65816

Joined:
Jan 4, 2005
#5
looks like others are helping you out pretty well in exampleing it. Algebra is kind of one of those things that kind of just clicks. It hard until you get to that clicking point and then you can do it in your sleep. That or any one who been though at least Cal 1 is really good at it. Calcus is 90% algebra and 10% calculas (heck all my up level math class is like 90% or more just algebra and very little using the therois and stuff.

That being said algebra is one of those math couse that you will use though out your entire life and is useful all the time. Up leval stuff you may not see it so much but Algreabra is used almost daily.

That being said stick with it and you will get it. Like others here I can help you out pretty easily. Just algebra is some time hard for me to help others out on because it is so easy for me. Curse of teh engineering mind. Hard to explain the more basic stuff at times.

Also as a bright spot when you hit the upper level stuff algebrea will be pretty easy and you also can become lazy and just go though the calculator. I know how to do it but a lot of the times I just put it in my cal because I dont want to make a stupid mistake completley friing 20-30 mins worth or work on a minor step.

6. ### rdf8585 thread starter macrumors regular

Joined:
Feb 15, 2006
#6
This is my last Algebra class, thank God. All I have left after this is Statistics, probably an intro-type level courses. Everyone seems to think it's easier than Algebra, and that I'd do much better in it. Who knows, anyone here have experience with statistics vs algebra?

7. ### 2nyRiggz macrumors 603

Joined:
Aug 20, 2005
Location:
Thank you Jah...I'm so Blessed
#7
Statistics I love....never really cared for algebra...but don't sleep on Stats...it aint easy.

Bless

8. ### rdf8585 thread starter macrumors regular

Joined:
Feb 15, 2006
#8
For this one I've got two answers

(2x^-1y^2z^3)^-3 multiplied by (3x^2y^-3z^-4)^2 answer should contain only positive exponents

I first got 2y^8 z^11 - 27x^5 over xy^6 z^8

then I got 9/(8 y^2 z^2)

Are either of those right?

9. ### Timepass macrumors 65816

Joined:
Jan 4, 2005
#9

You will be using algebra in those classes. Almost everything builds ontop of algebra. Algebra is the base.

Joined:
Apr 24, 2003
Location:
Colly-fornia
11. ### jsw Moderator emeritus

Joined:
Mar 16, 2004
Location:
Andover, MA
#11
If I correctly interpret the original equation as:

(3 x^2 y^-3 z^-4)^2
------------------------
(2 x^-1 y^2 z^3)^3

Moving things to the proper side [(y^-3 z^-4)^2 to the denominator as (y^3 z^4)^2, (x^-1)^3 to the numerator as x^3] yields:

(3 x^2)^2 * x^3
--------------------------
(y^3 z^4)^2 * (2 y^2 z^3)^3

or

9 x^7
--------------
8 y^12 z^17

If, of course, I'm interpreting it correctly.

12. ### Spanky Deluxe macrumors 601

Joined:
Mar 17, 2005
Location:
London, UK
#12
Algebra is a basic form of advanced mathematics.

13. ### _Matt macrumors 6502

Joined:
Aug 24, 2005
#13
do you know how to factor polynomials? for the second one thats what u need to do...

(x^2 - x) / (x^2 - 1) divided by (x^2 - 2x) / ( x^2 + 3x + 2)

(x(x - 1)) / ((x + 1)(x - 1)) multiplied by (keep, change, flip so...) ((x + 2) (x +1)) / (x(x - 2))

so hmm... the pairs of x, (x - 1), and (x + 1) cancel...leaving you with a final answer of

(x + 2) / (x - 2)

sry if im a horrible teacher but im pretty sure thats right...

14. ### bikegrrrl macrumors newbie

Joined:
Jul 6, 2006
#14
For the first problem you need to expand it out like you're multiplying binomials together (FOIL). Make sure you multiply the radicals and integers together correctly (radicals with radicals, integers with integers) and then add like terms together at the end.

For the second problem, instead of dividing, multiply by the reciprocal. All four of those algebraic expressions will factor, so do that and then reduce out any factors that are exactly the same. Then rewrite whatever is left into a new expression.

For the third problem, there are a bunch of ways to do it, depending on how you like to do them. For me, I would first get rid of the parenthesis by raising everything inside to the correct power. (when raising a power to a power, multiply the powers together) Then after you've done that, combine the left terms with the right. (when multiplying expressions with the same base, add the powers together). Lastly, get rid of negative exponents by moving them to the denominator. (x^-a = 1/(x^a))

Good Luck and hope this helps!

Cindy

15. ### Coolnat2004 macrumors 6502

Joined:
Jan 12, 2005
#15
Oh jeeze..

You guys reminded me that I have a "summer packet" for Pre-cal that I have to do.

I think I'll wait another month.

16. ### rdf8585 thread starter macrumors regular

Joined:
Feb 15, 2006
#16
These are some others, I actually did them, but can anyone verify I did them right?

a) 2 / (3*sqrt(5)) = 2sqrt5 / 15

b) 2 / cuberoot(2) = cbrt 4

c) 5 / (1 - sqrt(3)) = 5+5sqrt3 / 2

d) 3x^3-3x+7 / x-2 = 3(x^2+2x+3) + 25/(x-2)

17. ### Timepass macrumors 65816

Joined:
Jan 4, 2005
#17

umm you cannt do that with power and raising everything inside to the 3rd power. only workings with mutiplaying.

'

but if I read you 3rd problem right (2*x^.1*y^2*z63)^-3*(3x^2*y^-3*z^-4)^2

then the correct answer is (9*x^7)/(8*y^(23/4)*z^17) I kind of just ran it though my calculator and told it to expaind it out.

Some how I think you are writing down the 3rd problem incorrectly.

18. ### katie ta achoo macrumors G3

Joined:
May 2, 2005
#18
Oh man.. and I have a calculus placement exam soon...

Good luck with all this.. if worst comes to worst, just open up grapher and graph it.

that's how I got through algebra... Graph everything.

19. ### Electro Funk macrumors 65816

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Dec 8, 2005
Location:
The Opium Garden
#19
stats 1 & 2 were may favorite college courses...

20. ### Electro Funk macrumors 65816

Joined:
Dec 8, 2005
Location:
The Opium Garden
#20

2 / (3 * sqrt(5)) = 0.298142397

2 / cube root(2) = 1.58740105

5 / (1 - sqrt(3)) = -6.83012702

i have no idea on the last one neither does google calc...

21. ### Atlasland macrumors 6502

Joined:
Aug 20, 2005
Location:
London, UK
#21
The answer to number 1 isn't "2". You should get:

34 - 24*sqrt(2)

Number 2 is correct. As for number 3, the original problem is written so badly, that I really have no idea what the question is supposed to be!

22. ### rdf8585 thread starter macrumors regular

Joined:
Feb 15, 2006
#22
Number 3 came from my prof ....... i couldnt make much sense out of it either

23. ### Atlasland macrumors 6502

Joined:
Aug 20, 2005
Location:
London, UK
#23
A and B are correct.

You could write B as 2^(2/3) instead of cbrt 4. But cbrt 4 is correct.

C should be -(5+5sqrt3 / 2) instead of (5+5sqrt3 / 2). [you missed a negative sign on the denominator]

D: I'm not entirely sure what is wanted here. The original expression is in a "simpler" form than your answer. Moreover, I don't think that the original expression can be simplified easily, because 7 is prime.

24. ### bikegrrrl macrumors newbie

Joined:
Jul 6, 2006
#24
a, b, and d are correct. For c, when you multiply by the conjugate, 1+rad3, your denominator should come out -2.

personally, I wouldn't factor out the answer for d and I would leave it 3x^2 + 6x + 9 + 25/(x-2)

25. ### ToddW macrumors 6502a

Joined:
Feb 26, 2004
#25
yep forgot to use the FOIL method, having no variable threw me off. I barely glanced at that one, should have thought about it a bit more. and i wasn't sure how to do number 3 just going on the whole positive exponent.