ok well, im wokring on an assignment, and its asked.

The sum of a number and it's reciprocal is 30. Find the number.

please _dont_ do it for me but can someone show me how to work out the answer with say a different number.

thanks.

edit: dont' do it for me

quote:

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Originally posted by Ash_
The sum of a number and it's reciprocal is 30. Find the number.

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rofl yeh fractions are ok.

that crap looks like a different language.

i obviously wasnt paying attention when he showed us the example .


*is thinking about deleting the Elisha pics on his phone

no how is that bragging?
i dont understand what JA wrote it looks really complicated. is it supposed to be a joke cause it is scaring me that i dont know how to do it at all?

hold on. i took what you said the wrong way, i thought you were saying that i was stupid because i didnt understand it. sorry lol.

umm on MSN he told me its his formula. so maybe thats why it looks so complex because its like a custom one ?

sorry again for the misunderstanding

Ya i was just helping him with maths. If i can help then why not ...
What i said was this.
let the no be "x"
The no+its reciprocal is x+1/x ... and that should be 30
So x+1/x=30
Multiply throughout by x
x^2+1=30x
Note:- x^2 means x raised to the power 2 or x square
so the eq is x^2 - 30x + 1=0
Now we need to factorise this into 2 numbers such that the product is 1 and the sum is 30
WTF thats not possible
The sum is wrong

quote:

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Originally posted by Ash_
i obviously wasnt paying attention when he showed us the example .

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Well, I hate to repeat stuff, but maybe this looks a bit less like 'Chinees' (no punn intended to the Chinees people ):

Given:
  x + reciprocal(x) = 30
  (where reciprocal(x) = 1/x)

Search: x

Solution:
  x + reciprocal(x) = 30
  => x + 1/x = 30
  => (x + 1/x) * x = 30 * x
  => x² + 1 = 30 * x
  => x² - 30x + 1 = 0

this is a normal quadratic equation in the form of:
  ax² + bx + c = 0     this formula is basic knowledge!
so:
  a = 1   ,   b = -30   ,   c = 1

and to find x:

     this formula is basic knowledge!

(mind the ± in front of the square root, this means there are two possible solutions)

^^ now solve that normal equation:

the discriminate (b² - 4ac):
  => b² - 4ac = (-30)² - 4 * 1 * 1
  => b² - 4ac = 896
the discriminate is positive so we will get a 'real' number. (dunno if you already studied 'complex' numbers or also called 'imaginary' numbers. But if the discriminate is negative you will take the square root out of a negative number, resulting in a complex (or imaginary) solution for x.

now to go further with calculating x:
  => x = [ 30 ± sqr(896) ] / 2
  => x = 30 / 2 ± sqr(896) / 2
  => x = 15 ± sqr(896) / sqr(4)
  => x = 15 ± sqr(896 / 4)
  => x = 15 ± sqr(224)

solution 1:  x = 15 + sqr(224)
solution 2:  x = 15 - sqr(224)


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PS: John Anderton, you made a big mistake; you forgot the square root in your formula!
"Use [-B +- (B^2 - 4 AC)^1/2] / 2A"

quote:

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Originally posted by CookieRevised

PS: John Anderton, you made a big mistake; you forgot the square root in your formula!
"Use [-B +- (B^2 - 4 AC)^1/2] / 2A"

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quote:

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Originally posted by Hobbes

whats a reciprocol? Is it like the reciprocol of 2 is 1/2?

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quote:

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Originally posted by John Anderton

quote:

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Originally posted by CookieRevised
PS: John Anderton, you made a big mistake; you forgot the square root in your formula!
"Use [-B +- (B^2 - 4 AC)^1/2] / 2A"

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That was a typo ...

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quote:

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Originally posted by John Anderton
and i told u the eqn doesnt have any real roots only such odd irrational ... (4go what they r called; they aint irrational) numbers.

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Cookie is right. I just started studying this equation with 2 answers.

any problem that is like

ax^2 + bx + c  = 0

you can find two answers for. There is more then one equation to do so. But the one Cookie posted is a better one for this equation.

quote:

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Originally posted by CookieRevised
a massive typo if I made add

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quote:

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Originally posted by CookieRevised
(and what you mean are complex numbers or imaginary numbers)

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quote:

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Originally posted by Hobbes
Cookie is right. I just started studying this equation with 2 answers.

any problem that is like

ax^2 + bx + c  = 0

you can find two answers for. There is more then one equation to do so. But the one Cookie posted is a better one for this equation.

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quote:

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Originally posted by John Anderton
and i told u the eqn doesnt have any real roots only such odd irrational ... (4go what they r called; they aint irrational) numbers.

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quote:

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Originally posted by John Anderton
He said it in detail ... i just multiplied and showed it cause i was in a hurry ... i didnt show the answer as it wasnt a perfect whole number I told all this on msn

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quote:

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Originally posted by CookieRevised
(4go what they r called; they aint irrational) numbers.

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got it here

quote:

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Originally posted by CookieRevised
tbh, I don't care what is said on msn. This is a forum, everybody can read this, so either explain it correctly (especially in such math cases) or don't

(this whole thread could be as small as a post or 3 when things were explained correctly and detailed in the first place)

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* John Anderton cries and runs as he is no match for the Cookie's reply to series

* John Anderton is a cookie fan btw

quote:

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Originally posted by CookieRevised

But this is not what people mean when they talk about imaginary numbers. Imaginary numbers are numbers that can be written as a real number times i, where i=sqr(-2)

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I already posted the proper and simple definitions of the namings of numbers 15 minutes ago... (and moved them to my previous post)... So reload the thread before answering...

quote:

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Originally posted by John Anderton
i = sqr (-1) .....

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oh ok i think i see what you are doing a bit more i have used that equation. and dont worry about the misunderstanding Ash i do it all the time

Ok i have a maths problem .... Its a sum of probability .... I got the answer but probability is a bit dodgy and its so easy to make a mistake .... Can some one also try this problem

There is a dictionary containing only 5 letter words and there are only 4 letters viz. a, b, c, d, e.
The 1st word is aaaaa the second is aaaab the third is aaaac and so on.
Now we need to find the position of the word "adbca"

The answer i worked out is 216 .... can someone back up my answer ...... none of my friends have solved it ..... plus group solving aint allowed (if caught then im screwed )

a   b   c   d   e
0   1   2   3   4

adbca
03120

(0*5^4)+(3*5^3)+(1*5^2)+(2*5^1)+(0*5^0)
(0*625)+(3*125)+(1*25)+(2*5)+(0*1)
0+375+25+10+0
410

410/5 82 R 0
82/5 16 R 2
16/5 3 R 1
3/5 0 R 3

3120

quote:

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Originally posted by CookieRevised
Well, I hate to repeat stuff, but maybe this looks a bit less like 'Chinees' (no punn intended to the Chinees people ):

Given:
  x + reciprocal(x) = 30
  (where reciprocal(x) = 1/x)

Search: x

Solution:
  x + reciprocal(x) = 30
  => x + 1/x = 30
  => (x + 1/x) * x = 30 * x
  => x² + 1 = 30 * x
  => x² - 30x + 1 = 0

this is a normal quadratic equation in the form of:
  ax² + bx + c = 0     this formula is basic knowledge!
so:
  a = 1   ,   b = -30   ,   c = 1

and to find x:

     this formula is basic knowledge!

(mind the ± in front of the square root, this means there are two possible solutions)

^^ now solve that normal equation:

the discriminate (b² - 4ac):
  => b² - 4ac = (-30)² - 4 * 1 * 1
  => b² - 4ac = 896
the discriminate is positive so we will get a 'real' number. (dunno if you already studied 'complex' numbers or also called 'imaginary' numbers. But if the discriminate is negative you will take the square root out of a negative number, resulting in a complex (or imaginary) solution for x.

now to go further with calculating x:
  => x = [ 30 ± sqr(896) ] / 2
  => x = 30 / 2 ± sqr(896) / 2
  => x = 15 ± sqr(896) / sqr(4)
  => x = 15 ± sqr(896 / 4)
  => x = 15 ± sqr(224)

solution 1:  x = 15 + sqr(224)
solution 2:  x = 15 - sqr(224)


-------

PS: John Anderton, you made a big mistake; you forgot the square root in your formula!
"Use [-B +- (B^2 - 4 AC)^1/2] / 2A"

---

quote:

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Originally posted by John Anderton
Ok i have a maths problem .... Its a sum of probability .... I got the answer but probability is a bit dodgy and its so easy to make a mistake .... Can some one also try this problem

There is a dictionary containing only 5 letter words and there are only 4 letters viz. a, b, c, d, e. The 1st word is aaaaa the second is aaaab the third is aaaac and so on. Now we need to find the position of the word "adbca"

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quote:

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Originally posted by John Anderton
(...) and there are only 4 letters viz. a, b, c, d, e.

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quote:

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Originally posted by John Anderton
The answer i worked out is 216 .... can someone back up my answer

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quote:

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Originally posted by Chrono
oh thakns cookie, yu are so brilliant
now help me please with this problem:



(its not hard, but i would love to see how does he manage to write the solution here )

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sorry that should have been 4 letters .... a,b,c,d and e
viz. = namely ....
Its not probability .... its using permutation and combination

quote:

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Originally posted by CookieRevised
To calculate the equivalent base ten number:
= 0 * 5^4   +   3 * 5^3   +   1 * 5^2   +   2 * 5^1   +   0 * 5^0
= 410

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quote:

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Originally posted by John Anderton
The answer i worked out is 216

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quote:

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Originally posted by John Anderton
Where did i go wrong

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quote:

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Originally posted by CookieRevised
This is not yet the answer! Now it is time to read the question again. They asked what position the number is in. And positions useually start at 1. To make no confusion say this also in your answer :

= 411 (with first position being 1)

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quote:

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Originally posted by John Anderton
sorry that should have been 4 letters .... a,b,c,d and e

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quote:

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Originally posted by John Anderton
I dunno that formula that u r using .... can u do the same question with permutation and combination .....

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quote:

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Originally posted by John Anderton
1st word aaaaa
2nd        aaaab
3           aaaac
4           aaaad
5           aaaba

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quote:

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Originally posted by John Anderton
Ok so 4 changes r possible in the 1st letter (from the right) (like units place in numbers) ....

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quote:

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Originally posted by John Anderton
5th change (note: this is wrong if there are only 4 possible letters... it's the 4th change!!! see above) will make the 2nd letter from the right change once. So When the 1st letter changes 25 times (5*5) then the 3rd letter from the right will change. When the 1st letter changes 125 times then the 4th letter changes once .....
the 4th letter from the right changes 3 times (a -> b -> c ->d) thus ..... 125*3 = 375

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quote:

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Originally posted by John Anderton
so a total of 410 words have passed and we have reached adbca. So shouldnt that be the answer

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quote:

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Originally posted by CookieRevised
(provided there are 5 possible letters (a, b, c, d, e)! Which is apparently not what you've showed in your last dictionary in your last post, so make up your mind on the number of possible letters.)

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the answer is:

- total numbers = 4 possible letters: a, b, c, d
- number of changes (value of the letter) for unit 0 in "adbca" = 0  => 0 * 4 ^ 0
- number of changes (value of the letter) for unit 1 in "adbca" = 2  => 2 * 4 ^ 1
- number of changes (value of the letter) for unit 2 in "adbca" = 1  => 1 * 4 ^ 2
- number of changes (value of the letter) for unit 3 in "adbca" = 3  => 3 * 4 ^ 3
- number of changes (value of the letter) for unit 4 in "adbca" = 0  => 0 * 4 ^ 4
- 0*4^4 + 3*4^3 + 1*4^2 + 2*4^1 + 0*4^0 = 216

=> position or row number 217

Cookie, explain how you can have the square root of minus 1?!? I mean, I thought/think that the square root is like the opposite of ^2, as + is opposite to -. Since x^2 = x*x, the answer is always positive ( -3 * -3 = 9 and 3 * 3 = 9). So you can't have [something]^2 and then get -1, so you can't do sqrt(-1) either. That's what I learned anyway.. Dutch schools (or Cookie)

Excellent and interesting question though. But, you will learn that in the futur. It's the same as you didn't learn about integrals and deviations yet. Everything will come in time, and you first need to understand other things before you can understand this

Anyways, here is a shot (and some interesting history of maths)...

No, taking a square root of a negative number is, as you pointed out, impossible in the real world. However, that doesn't mean you can't do maths with the "impossible". It's something like the number "infinity". Infinity doesn't exist in real life, everything is limited. But you still need something to do maths with and to represent "something" beyond our scope of thinking.

This is the same for taking a square root of a negative number. In some calculations like quadratic and cubic equations (like used ealier in this thread) you will see that you can have a result which actually can't exist, but still it _is_ a result.

In the past, people stopped there, they couldn't go further. But one day, somebody said, what if we just invent a new number, and lets call it i (from "imaginary number"), i would be equal to the square root to -1. From that time on people could do maths and work out solutions further where they were stuck in the past.

note: such things happened in the past also (and still are happening now). It is only since the 13th century that negative numbers were slowly considered real numbers in the Western world. Before that, anything lower then 0 simpy didn't exist.

So, a whole new group of numbers emerged, the "complex numbers" written as "(x, y)" where x is a real number and y the imaginary number. So the complex number (5, 4) means 5+4i or 5+4*SQR(-1)...

So you can do maths with them too of course:
(5, 4) + (5, 4) = 5+4i + 5+4i = 10 + 8i

And you can even solve this strange question: divide the number 10 into two parts so that their product is 40! Impossible you would think, well not with complex numbers, the solution: 5+SQR(–15) and 5–SQR(–15). The proof:

[5+SQR(–15)]  +  [5–SQR(–15)]
=>  5 + 5 + SQR(–15) - SQR(–15)
=>  5 + 5
=>  10, hey presto!

[5+SQR(–15)]  *  [5–SQR(–15)]
=>  5²  -  5*SQR(–15)  +  5*SQR(–15)  -  SQR(–15)²
=>  25  -  -15
=>  25  +  15
=>  40, hey presto!

Note: this is actually a "famous" calculation/question from Cardano (16th century). He didn't actualy invented the imaginary number i and also didn't do anything further with this calculation or understood what he just found. This was actually the result of a competition to solve the cubic equation. Which was thought to be impossible. Cardona solved it by using negative numbers (a great controversy in that time) and this "complex" number thing was only a small byproduct of it.

Cardano did not go further into this, and it is only later that they were called complex numbers. A few years later Bombelli gave several examples involving these new type of numbers.

eg:

One of Cardano's cubic formulas gives the solution to the equation:
x³ = cx + d
as
x = SQR³[d/2 + SQR(e)] + SQR³[d/2 – SQR(e)]    where e = (d/2)² – (c/3)³
I've used "SQR³(x)" to indicate the third root of x

The 'problem' is that e could be negative, but there isn't any real number which you can multipy by itself and get a negative number, so Cardano was stuck there. Nevertheless Bombelli used this formula to solve the equation:
x³ = 15x + 4

according to Cardano's formula:
c = 15
d = 4
thus e = (4/2)² – (15/3)³  =>  2² - 5³  =>  -121  (a negative number no less!!!)
and thus x = SQR³[2 + SQR(–121)] + SQR³[2 – SQR(–121)]

Now, the square root of –121 is not a real number; it's neither positive, negative, nor zero. Nevertheless, Bombelli continued to work with this expression until he found equations that lead him to the solution 4, a real number!
So there _is_ a real solution eventhough SQR(–121) can't exist.

This shows nicely how complex numbers do have their use to solve things, even though they can't exist. Prior to the discovery of this, people said there was no solution. Although 4³ is equal to 15 * 4 + 4....


There is more detail to this story though:

When cubic formulas were a big hush-hush in the 16th century, Cardano noted that the sum of the three solutions to a cubic equation (x³ + bx² + cx + d = 0) is –b, the negation of the coefficient of x².

By the 17th century the theory of equations had developed so far so that someone called Girard developped a principle of algebra out of this. He also said that an nth degree equation always had n solutions. This is what we now call "the fundamental theorem of algebra". But He never was able to proof this principle though, just because some solutions couldn't "exist".

Another guy also studied this relation between solutions and coefficients, and was able to show that Cardano and Girard were right if you disregarded negative solutions as "false" and those other impossible solutions as "imaginary".

And slowly negative numbers became a common good and raised in status (by the results and efforts of other people), but it was not until the 18th century that complex numbers became in real use, roughly 200 years after their first "discovery" . They weren't considered to be real numbers though, but they were useful in the theories and formulas back in that time.

But still nobody was able to proof the fundamental theorem of algebra which stated that there are n solutions of an nth degree equation. So, although the complex numbers were not fully understood, the square root of -1 was being used more and more.

By the end of the 18th century numbers of the form x + yi were used pretty often by research mathematicians. For example Euler, a very influencial mathematician used it often in his trigonometric functions. And because of this and his influences, it was Gauss, and others like Wessel and Argand, who began to study complex numbers in function of xy-planes (as coordinates of planes) and that complex numbers were fully understood. And thus in 1799, Gauss published the first proof to the Girard's principle of algebra.

(You might know Gauss from the "Gauss-curve", it's the same one. And Euler is also a name which you certainly will hear in the next years in school, if you already didn't. And this whole deal with complex numbers will certainly be on the agenda and you'll notice that it will all start with talk about coördinates of planes and the polar coördinates system )

holy shit...



im suuuuuuuch an asshole for not understanding a  thing of this ! *cries*

a quote from one of my maths proffessors...

"life is complex....
...it has real and imaginary parts"

quote:

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Originally posted by Fredzz
im suuuuuuuch an asshole for not understanding a  thing of this ! *cries*

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quote:

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Originally posted by CookieRevised
- 0*4^4 + 3*4^3 + 1*4^2 + 2*4^1 + 0*4^0 = 216
=> position or row number 217

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Thanks aswell Cookie.

Just out of curiosity, what is the solution to the integral problem posted by cookie above? I hate calculus...

lol.

* Dane found the answer

I hate Math. Evidently Plus! does too .  Ask a smart person like wtbw or Cookie or Patchou or Time or Matty or Someone else I havent listed !