1. ## (1000/205-x)-(1000/205+x)=1

Hello, you guys can call me Azusa.; I am new to this forum. Sorry to introduce myself in such an impolite way, but I have a test at 8am later (and yes I'm pulling an all-nighter to review everything). I've focused on one area of the chapters so much that I have no idea how to do the other problems.

I appreciate any help!!

Thank You,
Azusa.

(One of the problems on the Quadratic Equations chapter).

2. ## Re: (1000/205-x)-(1000/205+x)=1

Multiply each term (both sides) by (205-x)(205+x) Get 1000(205+x)-1000(205-x)=(205-x)(205+x)

So 205000+1000x-205000+1000x= 42025-x^2 So x^2+2000x-42025=0 Formula gives x= 20.8 or -2020.8

3. ## Re: (1000/205-x)-(1000/205+x)=1

Ok, I got to x²+2000x-42025=0
but I don't quite get how you got x=20.8 or -2020.8

After getting x²+2000x-42025=0, I used the Quadratic Forumla; (-b±√b²-4ac)/2a
Ended up with this: x=(-2000±√3831900)/2
Then √3831900 (calculator): 1957.523946, thus, x=(-2000±1957.523946)/2

I don't get how to do that last step.. Am I supposed to add/substract -2000 and 1957.523946 then divide by 2? but that doesn't seem to get me the answer..

Thank You,
Azusa.

4. ## Re: (1000/205-x)-(1000/205+x)=1

Your number under the square root sign is wrong. We have a=1 b=2000 and c=-42025

So b^2-4ac= (2000)^2-4(1)(-42025) =4000000+168100=4168100

5. ## Re: (1000/205-x)-(1000/205+x)=1

Wow. x.x What a dumb mistake. I just wasted 3 hours being stupid. :c
Anyway, I APPRECIATE your help. Couldn't thank you enough.

TYVM,
Azusa.