Shoutbox

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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x+1/x=30
x^2 + 1 = 30x
X^2 - 30x +1 = 0
A x^2 + B x + C = 0
Is a fraction ok ?? Cause its gotta be
Use [-B +- (B^2 - 4 AC)1/2] / 2A = Roots
ie roots are [-B + (B^2 - 4 AC)^1/2] / 2A And [-B + (B^2 - 4 AC)1/2] / 2A
Ill work it out

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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The example must be different cause i was dwelling on it while i was playing soccer for the past hour or so and ... not possible
U need to get the question again Call someone up

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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That was a typo ... and i told u the eqn doesnt have any real roots only such odd irrational ... (4go what they r called; they aint irrational) numbers.

EDIT :-

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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Reciprocal is any number taken as 1/
Yes the reicprocal of 2 is 1/2 and vice versa ....
Divide the number by 1 and it becomes (is called actually) its reciprocal.

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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a massive typo if I made add

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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the solution does have 2 real numbers...
(and what you mean are complex numbers or imaginary numbers)

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.