- **Algebra**
(*http://mymathforum.com/algebra/*)

- - **number and sets**
(*http://mymathforum.com/algebra/18075-number-sets.html*)

number and sets1.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. |

Re: number and sets... x = 4 is out. |

Re: number and sets1.) 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: |

Re: number and setsAccording 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. |

Re: number and sets |

All times are GMT -8. The time now is 12:37 AM. |

Copyright © 2019 My Math Forum. All rights reserved.