OK, math problem

[phoenix_nf]

Ok.

i'll post it in a new thread...the math equation i put forth was not related to the problem at hand. it was just a neat equation to break your head on.

[sven]

ti-92s response: non factorable (this is a factoring calculator lol)

[XtremeMaC]

mathlab? mathematica anyone?

[D.Celt]

Take square roots on line one and you get the same error.

[studguy1]

phoenix:

The error is in the square root step

(4-9/2)^2 = (5-9/2)^2

Taking Square root

4-9/2 = -(5-9/2) (Negative root for R.H.S )

-0.5 = -0.5

THe Universe is safe again

Matrix:

I dont think the equation can be simplified anymore. It is already in its simplest form

IF you need to factorize it, it can be expressed as

(x+1.74296) *(x - 0.37148 + 1.46866 i) * (x - 0.37148 - 1.46866 i)

where each of the terms next to x is the root of the eqution. As stated before 2 of them are imaginary

[phoenix_nf]

studyguy1 you really study a lot

[MadGutts]

OMG !

When i was a t shool it was called math... then college called it Quatitatitive methods! then uni i gave up !!! LOL

scared off by silly, useless eqations that you need a super computer to calculate...

I'll stick with the hardware on pc's i think..... Less to go wrong ! lol

[Zxian]

For those of you who have studied complex numbers, leave this one alone...

For those of you who haven't... bust your brains on this...

if:

3^2 = 3*3 = 9

and

i^2 = -1

then what is:

i^i = ?

and

log(i) = ?

As for the second question, a good hint is to use Euler's Equation

e^(i*pi) + 1 = 0

and figure it out from there.

Cheers!

P.S. Don't just use your calculator... numbers mean nothing in math, method is everything.

[Spyder2k]

i^2 = -1

I must have missed something in school because it isnt possible to square a number and get a negative answer.

[sonu27]

You need to know what x is to solve it.

[zprog]

i^2 = -1

I must have missed something in school because it isnt possible to square a number and get a negative answer.

Oh, man. I think you did.

Anyway. Given Euler's formula you have to define log(i) in terms of i, I've arrived to:

log(i) = (pi*i)/(ln 100)

If you play around with Euler's formula (Note that log(-1) = pi * i), then you should get something like:

i^i = e^(-pi/2)

[Spyder2k]

i^2 = -1

I must have missed something in school because it isnt possible to square a number and get a negative answer.

Oh, man. I think you did.

You just nullified my high school math life

Could you give me an example of one such number, or point me in the direction where I can learn more about this?

[zprog]

The number i is essentially the square root of -1. So, for example, what is the square root of 4? It is 2. What is the square root of -4?

sqrt(-4) = sqrt(-1 * 4) = sqrt(-1) * sqrt(4) = sqrt(-1) * 2. Since sqrt(-1) = i, sqrt(-4) = 2i.

Then you get to do cool differential equations to figure out that e^(pi * i) = -1

Then, all you do is simple algebra to arrive to what i^i and log(i) are.

Here is a link: http://mathforum.org/dr.math/faq/faq.imag.num.html

[mark]

The number i is essentially the square root of -1. So, for example, what is the square root of 4? It is 2. What is the square root of -4?

sqrt(-4) = sqrt(-1 * 4) = sqrt(-1) * sqrt(4) = sqrt(-1) * 2. Since sqrt(-1) = i, sqrt(-4) = 2i.

Then you get to do cool differential equations to figure out that e^(pi * i) = -1

Then, all you do is simple algebra to arrive to what i^i and log(i) are.

Here is a link: http://mathforum.org/dr.math/faq/faq.imag.num.html

i is the imaginary square root of -1 ?

[Spyder2k]

Seems this technique is credible, I'll accept it but it seems the sole purpose of it's creation was for a want of a square root of negative numbers.

i would really like einstein to take a look at this because it seems not only that

sqrt(negative #) = i but sqrt(negative #) = -i

neway my IQ can only go so high

If this is the case, then i is not a real number and the solution for "e^(i*pi) + 1 = 0" would have to be in terms of i.