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Mar 19th, 2002, 08:30 PM
#1
Thread Starter
Fanatic Member
Please solve
I have a question, can anyone solve this equation step by step and explain?
Code:
I/{sqrt[(x-10)^2+(y-0)^2]}^2=2I/{sqrt[(x-20)^2+(y-0)^2]}^2
If you want you can turn to page 578 on Algebra 2 book, it mentioned there, I do not understand how after simplify, it can form x^2+y^2=200
Thanks

prog_tom
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Mar 19th, 2002, 10:10 PM
#2
Frenzied Member
Re: Please solve
Originally posted by prog_tom
If you want you can turn to page 578 on Algebra 2 book
And I'm supposed to have your algebra book?
I'm bringing geeky back...
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Mar 19th, 2002, 10:17 PM
#3
Fanatic Member
Re: Re: Please solve
Originally posted by jpbtennisman
And I'm supposed to have your algebra book?
[clears throat] I beilive you are supposed to have algebra 2
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Mar 19th, 2002, 10:30 PM
#4
Frenzied Member
Re: Re: Re: Please solve
Originally posted by Gimlin
[clears throat] I beilive you are supposed to have algebra 2
haha, nice call
but still...
I'm bringing geeky back...
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Mar 20th, 2002, 05:35 PM
#5
Hyperactive Member
Re: Please solve
Originally posted by prog_tom
I have a question, can anyone solve this equation step by step and explain?
Code:
I/{sqrt[(x-10)^2+(y-0)^2]}^2=2I/{sqrt[(x-20)^2+(y-0)^2]}^2
If you want you can turn to page 578 on Algebra 2 book, it mentioned there, I do not understand how after simplify, it can form x^2+y^2=200
Thanks
I/{sqrt[(x-10)^2+(y-0)^2]}^2=2I/{sqrt[(x-20)^2+(y-0)^2]}^2
is the same as
I/[(x-10)^2+(y-0)^2] = 2I/[(x-20)^2+(y-0)^2]
or
I/[(x-10)^2+ y^2] = 2I/[(x-20)^2+ y^2]
Dividing both sides by I
1/[(x-10)^2+ y^2] = 2/[(x-20)^2+ y^2]
Cross multiplying both sides,
[(x-20)^2+ y^2] = 2 * [(x-10)^2+ y^2]
or
2 * [(x-10)^2+ y^2] = [(x-20)^2+ y^2]
Expanding powers
2 (x^2 + 100 - 20x + y^2) = x^2 + 400 - 40x + y^2
2x^2 + 200 - 40x + 2y^2 = x^2 + 400 - 40x + y^2
Moving the Right side terms to the Left
2x^2 + 200 - 40x + 2y^2 - (x^2 + 400 - 40x + y^2) = 0
2x^2 + 200 - 40x + 2y^2 - x^2 - 400 + 40x - y^2 = 0
2x^2 - x^2 + 200 - 400 - 40x + 40x + 2y^2 - y^2 = 0
x^2 - 200 + y^2 = 0
Taking the 200 to the RHS
x^2 + y^2 = 200
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Mar 20th, 2002, 07:40 PM
#6
Thread Starter
Fanatic Member
thanks, another question:
Please write sqrt(27) in terms of i

prog_tom
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Mar 20th, 2002, 09:46 PM
#7
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Mar 20th, 2002, 11:39 PM
#8
Thread Starter
Fanatic Member

prog_tom
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Mar 21st, 2002, 05:26 AM
#9
Originally posted by prog_tom
Youare that Moore dude?
Sorry, No. I'm just your friendly, neighborly Spider-Ma...oops!
Been reading the SpiderMan Book Adaptation, just kinda slipped out.
Actually, I'm not sure who you mean?
Anyways, like my Square root of 27 using i?
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Mar 21st, 2002, 03:52 PM
#10
Fanatic Member
Please write sqrt(27) in terms of i
What is i?
Do you mean that, if i = 3, sqrt(27) = i * sqrt(i)?
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Mar 21st, 2002, 08:51 PM
#11
Thread Starter
Fanatic Member
I guess in this case the i is referring to the complex numbers.

prog_tom
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