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hoyy1kolko March 14th, 2011 07:45 PM

number and sets
 
1.Given z1=2-i and z2 =4+3i.Express

in the form a +bi ,where and b are real numbers.

My solution

i just skip to the last two part


1+ 7/5i (is it correct?)

2.Solve the equation square root of (3x/(1-?x))=2

My solution

squaring the both sides

3x/1-?x=4
3x=4-4?x
9x^2+16z-6=0

Then i am stuck in here,please help.

The Chaz March 14th, 2011 07:54 PM

Re: number and sets
 








...
x = 4 is out.

MarkFL March 14th, 2011 08:03 PM

Re: number and sets
 
1.) and

I made the assumption that z_2-z_1 was the denominator of the 2nd term, you can use parentheses to make it clear. :mrgreen:


hoyy1kolko March 15th, 2011 06:41 AM

Re: number and sets
 
According to the solution of question 1 that you did,

how do you get 3/1+2i

from the part 6/z2-z1?

6/ 2+4i
3+3/2i (that's what I get ). Please tell me how you get 3/1+2i.

The Chaz March 15th, 2011 06:44 AM

Re: number and sets
 


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