Help Me Solve This problem

PapiChullo

PapiChullo

CAGiversary!
[(d-4)/d^2+2d-8] - [(d+2)/(d+4)(d-4)]

I guess it's supposed to be reduced down as much as possible. My gf's sister has it for homework, and I forget all the rules for that crap.
 
My calculator only carries it out so far, but here's what I've got:

stick+it-(in+her)pooper
 
[quote name='Scobie']My calculator only carries it out so far, but here's what I've got:

stick+it-(in+her)pooper[/quote]

According to my abacus .. he's correct. This does compute.
 
It doesn't really reduce into anything useful as far as I can see. Busywork for the loss!
 
[quote name='jmcc']It doesn't really reduce into anything useful as far as I can see. Busywork for the loss![/quote]

Isn't that what beginning math homework is all about?

TBW
 
[quote name='TheBlueWizard']Isn't that what beginning math homework is all about?

TBW[/QUOTE]This kind should be like Tetris or a Puyo Pop. You factor stuff and cancel until you get down to minimal sets of parentheses. I loved doing this shit when I was learning it, but the more you're left with the less satisfying it is.
 
[quote name='jmcc']It doesn't really reduce into anything useful as far as I can see. Busywork for the loss![/QUOTE]

Yeah, I'm thinking he's got a typo in there somewhere.
 
[quote name='PapiChullo'][(d-4)/d^2+2d-8] - [(d+2)/(d+4)(d-4)]

[/quote]
this is same as:

(d-4)/[(d-2)(d+4)] - (d+2)/[(d+4)(d-4)]

Need to get common denominator which will be:

(d-2)(d+4)(d-4)

Thus, need to multiple left part by (d-4)/(d-4) this is okay since that is same as multiplying by 1

Then need to multiply right part by (d-2)/(d-2)

This gets you:

[(d-4)(d-4)]/[(d-2)(d+4)(d-4)] - [(d+2)(d-2)]/[(d-2)(d+4)(d-4)]

This gives as a numerator:

(d^2-8d+16)-(d^2-4) which is -8d+20 #FIXED ERROR#

This gives as the demoninator:

(d-2)(d+4)(d-4) = (d-2)(d^2-16) = d^3-2d^2-16d+32

So you have:

[-8d+20] / [d^3-2d^2-16d+32]

Is that what she needed?
 
[quote name='jmcc']It doesn't really reduce into anything useful as far as I can see. Busywork for the loss![/QUOTE]

I don't think it's about reducing, I think it's about finding the 0's. If you multiply both parts by the LCD, which are all the factors of each denominator, each ratio's respective denominators cancel out leaving you to multiply each other by the opposite's denominator (essentially). After you do that, you add/subtract like terms and factor/complete the square/use the Quadratic formula depending on what you have to show.


Edit* theepi, Idon't think that's right. Once you multiply by the LCd, the denominators are cancelled out. There should be no denominator. I'm pretty sure it's one of those "Find possible 0's" problems.
 
Reality's Fringe;2660238 said:
I don't think it's about reducing, I think it's about finding the 0's. If you multiply both parts by the LCD, which are all the factors of each denominator, each ratio's respective denominators cancel out leaving you to multiply each other by the opposite's denominator (essentially). After you do that, you add/subtract like terms and factor/complete the square/use the Quadratic formula depending on what you have to show.


Edit* theepi, Idon't think that's right. Once you multiply by the LCd, the denominators are cancelled out. There should be no denominator. I'm pretty sure it's one of those "Find possible 0's" problems.
Click to expand...

You can't multiple by the LCD's and find zeroes, because it isn't an equation.
The directions were to reduce.
 
[quote name='theeipi'][(d-4)(d-4)]/[(d-2)(d+4)(d-4)] - [(d+2)(d-2)]/[(d-2)(d+4)(d-4)]

This gives as a numerator:

(d^2-16)-(d^2-4) which is -12[/QUOTE]I think you've got a typo. (d-4)(d-4) doesn't equal (d^2-16)
 
[quote name='theeipi']You can't multiple by the LCD's and find zeroes, because it isn't an equation.
The directions were to reduce.[/QUOTE]

Oh, pffft. I didn't see that part.
 
[quote name='PapiChullo'][(d-4)/d^2+2d-8] - [(d+2)/(d+4)(d-4)][/quote]

Just in case the problem was ABOVE = 0, solve for d:

multiple both sides by LCD:

Common Denominator was (d+4)(d-4)(d-2)

This gives:

(d-4)(d-4) - (d+2)(d-2) = 0

This gives:

d^2-8d+16 - (d^2-4) = 0

-8d +20 = 0

2.5 = d which is a not excluded from solution set...
but I don't think that was what was asked for here.
 
I tossed my scrap paper I did it on, but I think it was something like (-8d+20)/(d+4)(d-4)(d-2)
 
[quote name='jmcc']I tossed my scrap paper I did it on, but I think it was something like (-8d+20)/(d+4)(d-4)(d-2)[/QUOTE]

That is what I get to. Looks like the OP needs to try some polynomial division.
 
[quote name='UnderwaterMadman']isn't (d-4)(d-4) = d^2-8d+16?[/quote]

Only if you want to do it right!

Fixed above post. Wierd thing is, I did it right in below post.
 
[quote name='UnderwaterMadman']That is what I get to. Looks like the OP needs to try some polynomial division.[/QUOTE]I don't think you can do that with a higher degree polynomial as the divisor, though. I mean, you can, but it gets uglier than it already is.
 
[quote name='UnderwaterMadman']That is what I get to. Looks like the OP needs to try some polynomial division.[/quote]

Polynomial division only does something if the degree of the numerator isn't less than the degree of the denominator.

Otherwise, like here, you get 0 + Remainder the same as the numerator
 
[quote name='theeipi']Limit = 0, as it always does when Degree of Numerator < Degree of Denoninator[/QUOTE]

But do you know what happens when they're the same degree?! INSANITY! Dogs and cats living together, MASS HYSTERIA.


Also, maybe the coefficient ratio. Maybe.
 
Never was very good at taking down those math bosses. I could always get thru the first few levels pretty easy, but when I start getting surrounded by a bunch of parenthesis, I always panic and fire off all my "" and by the time the big math boss comes out, all I got are wimpy "^" and maybe a "%" or two.

I'm not totally sure, but I think the answer is "$#@%!!".
 
Ok, so they found out the correct answer, and it is.......





-8d+20/ (d-4)(d+4)(d-2)

Thanks for all the help. Cookie goes to those that came up with that answer.
 
[quote name='rajchakrabarti']aah crap.. this whole time i thought the answer was 3[/QUOTE]
That's funny, when I originally did it, I got like -3.

I forget this, so can someone explain how to get the LCD?
 
[quote name='PapiChullo']That's funny, when I originally did it, I got like -3.

I forget this, so can someone explain how to get the LCD?[/QUOTE]Look at both denominators and determine what each has that the other doesn't (usually involves factoring them completely). Then make them match by multiplying tops and bottoms by the missing terms.

So if you have x/y and -a/b you're missing b in the denominator on the left and y on the right, so x/y is multiplied by b/b (so it puts the b in the bottom, but doesn't really change anything) and the right by y/y. Then you have bx/by and -ay/by. Since the bottoms match, you can stick the tops together for (bx-ay)/by.
 
[quote name='theeipi']Thread of the Year Nominee![/QUOTE]

I don't know...that cat orgy thread is quite the competitor
 
[quote name='crazytalkx']I don't know...that cat orgy thread is quite the competitor[/quote]
That thread should be merged with this, based on obvious similarities.
 
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