Hey! I thought we were solving problems[mathematical] here.

My brothers i don't see insults as the objective of this thread.no one is above mistakes.we are all here to learn.it's obvious that some of us are graduates,some are still in school while others(like me) are still searching for admission so in that case we all must no be equall

...we need understandin 2 make progress...i suppose we 're all adults here...we should b a family...what i xpect 4rm maturd minds is cooperation...let's all learn 2 accept our mistaks, cos dats d surest way 2 upliftment....

Pls can someone helpme resolve this into partial fractions.....1/(x^3-1)

Pls can someone helpme resolve this into partial fractions.....1/(x^3-1).

...three numbers are in arithemetic progression, the product of any two when increased by one is a perfect square...find d numbers...

Cashio: Pls can someone helpme resolve this into partial fractions.....1/(x^3-1).

...1/(x^3 -1) =1/(x-1)(x^2 +x +1)....
..put 1/(x-1)(x^2 + x +1) = F/(x-1) + Gx+H/(x^2 +x +1)....=> 1=(Gx+H)(x-1) +F(x^2 +x+ 1)....should i continue frm here? Or u can finish it up..

Laplacian: ...three numbers are in arithemetic progression, the product of any two when increased by one is a perfect square...find d numbers...

is it the product of any two plus one or each of the two numbers plus one and their product?

Alpha Maximus: is it the product of any two plus one or each of the two numbers plus one and their product?

the product of any two plus one

doubleDx:

I agree, you can cover your face in shame; I see how low your IQ is, you can't even recognize a typo Trying to teach me English in Math class huh? Folks like you are not supposed to be on a Math thread! @ your age, you can't even apply the law of log properly? It's a shame ***Covering my face for you*** How did you pass WAEC math BTW?

chai bros, can't just believe all these are from you. You disappointed me big time, you are one of the nairalanders i really respect for being humble, can't you just abandon that mediocre, he is not ready to learn,
Imagine him requesting for your CGPA just because someone else corrected you, na too know they kill some folks sha.

Laplacian: ...three numbers are in arithemetic progression, the product of any two when increased by one is a perfect square...find d numbers...

-1,0,1

benbuks: ...1/(x^3 -1) =1/(x-1)(x^2 +x +1)....
..put 1/(x-1)(x^2 + x +1) = F/(x-1) + Gx+H/(x^2 +x +1)....=> 1=(Gx+H)(x-1) +F(x^2 +x+ 1)....should i continue frm here? Or u can finish it up..

i'll try..2maro

Laplacian: ...three numbers are in arithemetic progression, the product of any two when increased by one is a perfect square...find d numbers...

alright, here we go:
Let the first term be 'a' and the common difference be 'x'
1st term : a
2nd term : a+x
3rd term : a + 2x
Since the sum of the product of just two terms and 1 give the same answer, we have that:
(a(a+x))+1=(a(a+2x))+1=((a+x)(a+2x))+1
Therefore,
A^2 +ax+1=a^2+2ax+1=a^2+3ax+2x^2+1
Now there are three equations, we equate the first and last, and temporarily isolate the second(because we don't know which is equal to which yet)
a^2+3ax+2x^2+1=a^2+ax+1
Collecting like terms we have
2ax+2x^2=0
2x(a+x)=0
X=0(impossible,since its an arithmetic progression,there has to be a difference) and a+x=0, a=-x(correct)
Since 2ax+2x^2=0, the middle equation is also equal to 0
Thus,
a^2+2ax+1=0
Now substitute 'a=-x' into the middle/second equation
(-x)^2+2(-x)x+1=0
X^2-2x^2+1=0
Factorising, we have:
(X-1)(x-1)=0
X=1(twice)
Now that we have established the common difference of '1', we input it back into the original equations but this time we subtract 1 from all three sides for simplicity and we have:
a(a+1)=a(a+2)=(a+1)(a+2)
We first equate and solve the first and second
a(a+1)=a(a+2)
a^2+a=a^2+2a
Collecting like terms, we have
-a=0
Thus, a=0
Now we simply equate the third equation to 0
(a+1)(a+2)=0
Thus a=-1 and a=-2
We try out a=-2 as the first term of the series a, a+1, a+2 but it does not satisfy the question
But when a=-1 is tested, it satisfies the question(all products of a pair + 1=a perfect square)
Therefore the terms are -1, -1+1, -1+2=-1, 0 , 1
N.B: recall that in my solution , the middle equation was already equal to 0, before I arrived at my final three answers

benbuks: ...1/(x^3 -1) =1/(x-1)(x^2 +x +1)....
..put 1/(x-1)(x^2 + x +1) = F/(x-1) + Gx+H/(x^2 +x +1)....=> 1=(Gx+H)(x-1) +F(x^2 +x+ 1)....should i continue frm here? Or u can finish it up..

bros it's supposed to look like this 1/x^3-1 =1/x(x^2-1)
=A/x+B/(x^2-1)
1/x(x^2-1)=A/x+Bx+C/(x^2-1)
1/x(x^2-1)=A(x^2-1)+(Bx+C)x/x(x^2-2)
1=A(x^2-1)+(Bx+C)x
Let x b 0
1=-A
A=-1
To be continued Tomorrow a bit busy and tired

@ cashio questn.
resolve into partial fractn.
1/x^3-1.
solution.
1/(x-1)(x^2+x+1). Representing with A/(x-1) + Bx+C/(x^2+x+1). Nw we av
1=A/(x-1) + Bx+C/(x^2+x+1).
1=A(x^2+x+1)+(Bx+C)(x-1).
1=Ax^2+Ax+A+Bx^2-Bx+Cx-C.
1=x^2(A+B)+x(A-B+C)+A-C.
Equatin d coefficient
A+B=0... Eqn 1
A-B+C=0... Eqn 2
A-C=1... Eqn 3. Nw solvin simultaneously we av A=1/3, B=-1/3 nd C=-2/3. Putting back in A/(x-1) + Bx+C/(x^2+x+1). Finally we av.
1/3(x-1)+ (-x-2)/3(x^2+x+1).

Hmm.

@ pual paul my guy..x(x^2- 1)=x^3-x =/= (x^3 - 1)..i guess u r really..tayad...please you nid 2 rest
ok?.please verify b4 u argue...u try sha...
[/quote]

Calculusf(x):
...for(a)numbers which are divisible by 4 from 100-999...therefore it starts with 100 and has a common difference of 4 and last term will be 996(the closest to 999 which is divisible by 4) which is a.p...using the formula for last term of an a.p...nth=a+(n-1)d...where nth is the last term,a is the first term,n is the number of terms and d is the common difference...substitute those to the formula...996=100+(n-1)4...therefore 996-100=(n-1)4...896=(n-1)4...divide both sides by 4...224=n-1...therefore,make n the subject of formula...therefore n=225........(b)...the numbers divisible by 3 and 4 must start with 12 and multiple of 12...the first term closest to 100 which is a multiple of 12 is 108...and the last term is 996...the common difference is 12...then,substitute that to the formula for the last term of a.p....996=108+(n-1)12...996-108=(n-1)12...888= (n-1)12...divide both sides by 12...74=n-1...therefore n=75...

Pls develop the programme to determine it

Hidorano: 3. How many whole number from 100 to 999 are divisible by (a). 4 (b). both 3 and 4. God bless u all.

Who can develop a programme to determine that.

doubleDx:

With this crap & pride you would never learn anything....Are you saying the way you applied your log law was right? Man, you are a fuckin math kid...I don't argue with folks like you! mtchwww. I don't know what you think you know in math now....I dey laugh in tongues! Na una type dey claim say una write exams well but dem give una bad grades...yeye!

lol.

Alpha Maximus: ....haha!! Doubledx is a disappointment!! ....and mind you, I didn't request for his cgpa coz of corrections, I did it coz he said I was a math novice...get your facts right dude and mind your business@dekins88

Alpha Maximus, if you desist from posting here, we will understand. To tell you the truth, you still have a very long way to catch up in Mathematics. We know you have every right to comment wherever you want, but you see, with what I've seen, your commenting here is risky and you run the risk of infecting the unsuspecting Jambites among us with your wrong solution-right answers. Its not actually mostly your fault since you only teach and believe whatever was shoved down your throat by whom you called teachers, coupled with the fact that you dont know that the solutions are actually wrong.

I am saying this with lots of love. Cheers!

The five star general back online.
Thank God dat i have jst fully recovered from acute fever and fully ready to make contributions
I have gone through series of post by the math generals in the house and i noticed unhealthy verbal attacks between doubledx and alpha maxius. i want to believe that this thread is meant for matured math generals and there is no point throwing tandrums at our colleagues on the basis of correction. We are all here to learn from each other.

A@alpha maxiud,

Your solution to 2^x = 4x is wrong even if u get the answer right. What i am interested in is ur working. U show a total disregard to the basic law of logarithm.

Here is ur working:

2^x = 4x

Take log of both sides:

Xlog2 = log 4x
Xlog2=log4 +logx (since loga +logb=logab)
right up to this extend but here is where u get it all wrong and show a total disregard to basic law of logarithm

Xlog2 -log4=logx

Cancel out the logarithm of both sides:

2x-4=x
X=4. U have violated the law of basic mathematics and u are sentenced to 10years in imprisonment with hard labour by the math military police. Lol.

Simple rule of logarithm:

Loga +logb =logc
a+b is not equal to c

But logab=logc,

Only then, u can cancel out
ab=c.

Hope u get that.

Another common mistake again

Xloga=logb

xa is not equal to b

But loga^x=logb
a^x=b

So it's true that mathematicians are bounded with pride and anger.the battle was tough between general doubledx and recruit alpha maximus.lol

jackpot: Alpha Maximus, if you desist from posting here, we will understand. To tell you the truth, you still have a very long way to catch up in Mathematics. We know you have every right to comment wherever you want, but you see, with what I've seen, your commenting here is risky and you run the risk of infecting the unsuspecting Jambites among us with your wrong solution-right answers. Its not actually mostly your fault since you only teach and believe whatever was shoved down your throat by whom you called teachers, coupled with the fact that you dont know that the solutions are actually wrong.

I am saying this with lots of love. Cheers!

..my dear, i actually dnt kw what to tell him again,so i quit responding to his post, imagine i wast'nt even referin 2him..bt paul....please lets let him be...lets continue solving., learning frm each oda.nd helpin doz dat r ready 2 learn....

benbuks: ..my dear, i actually dnt kw what to tell him again,so i quit responding to his post, imagine i wast'nt even referin 2him..bt paul....please lets let him be...lets continue solving., learning frm each oda.nd helpin doz dat r ready 2 learn....

.i love that ben