[Guardsman]

Hey guys, I want to go to school for engineering, Electrical or Electronics (both, if they have a program like that).

Anyhow, I have a big problem that currently prevents me from finishing even my Associates (Community College of the Air Force).....I can't do text-book algebra.

I can do applied algebra, I've been doing it since I was a kid. But, put a text book in front of me, with x=axb+c2/d, and I might as well be reading Latin.

One test study guide that I looked at had a question asking what the square root of i is. I can't even begin to understand how I get an answer to that. My line of thinking is that an imaginary number doesn't exist, therefore, it can't have a square root. Nothing = nothing.

I worked at Stanford Linear Accelerator Center until recently, and I looked over some of the papers and machine technical stuff, in how the accelerator works, beam physics, etc.

When I look at that kind of stuff, it makes sense to me....I may not know the exact answer, but if I had all of the necessary information, I could figure out the answer.

I also picked up a book, "Engineering Formulas", by two German guys, and looking through there, the formulas make sense to me.

Has anybody had this kind of problem before, or know somebody that did, or just have any suggestions that might help me out?

I love doing math, but it frustrates the crap out of me that I can't get past this. I tried taking algebra twice in high school, flunked both times, tried it again in 1998, and got a D on that.

I don't want to just pass the classes, I actually want to learn the stuff. Any help would be appreciated. Thanks.

John

[pknowles]

Square root of i
i=(-1)^(1/2)
so to find the sqaure root of i you need to solve the following eqation:
(a+b*i)=i^(1/2) where a and b are what you need to find. a and bi are just the real and imaginary parts of some number whose sqaure is i. Note: a and b are real numbers (no i's hidden in them)
Sqaure both sides to get
(a+b*i)^2=i
expand the term (a+b*i)^2 to get a^2+2*a*b*i+(b^2)*(i^2)
now i^2=-1 and plug the expansion back into the formula to get
a^2+2*a*b*i-b^2=i
now sepperate real and imaginary parts
(a^2-b^2)+2*a*b*i=0+i
now you can split into 2 eqations to solve for a and b because the 2*a*b*i term is the only one with an i in it so 2*a*b=1 and a^2-b^2=0.
That's the hard part!
Now take the eqation a^2-b^2=0 and add b^2 to both sides to get:
a^2=b^2, so a=+b and a=-b
However if you take the next eqation 2*a*b=1 and plug in each answer (a=+b and a=-b)
if you plug in a=-b you get -2*b^2=1 which can't be satisfied if b is a real number, so a=b.
Now 2*b^2=1
divide both sides by 2:
b^2=(1/2) so b=1/((2)^(.5)) or the way math people like to see it (no sqaure roots in the denominator (sp?)) b=(2^(.5))/2 so plug those answers back into the very first eqation [(a+b*i)=i^(1/2)] at the top and [(2^(.5))/2+i*(2^(.5))/2=i^(1/2)]

DONE!

EDIT: -b) is a smily