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Subject: "Where do I start?"     Previous Topic | Next Topic
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Conferences The CTK Exchange High school Topic #178
Reading Topic #178
me
guest
Jun-04-02, 03:35 PM (EST)
 
"Where do I start?"
 
   4 (x+1)1/2 - 5(x+1)3/2 + (x+1)5/2 = 0


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bluediamond
Member since Apr-9-02
Jun-04-02, 09:10 AM (EST)
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1. "RE: Where do I start?"
In response to message #0
 
   I'm assuming you mean

4 * (x+1)1/2 - 5 * (x+1)3/2 + (x+1)5/2 = 0

In that case, x = -1 is obviously a solution. For other solutions, assume that x is not equal to -1 and divide the entire thing by (x+1)1/2 to get a quadratic equation.

dave


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Karl
guest
Jun-05-02, 09:10 AM (EST)
 
2. "RE: Where do I start?"
In response to message #0
 
  
First is that 4(x+1) * 1/2, that is multiplied by a half, or 4(x+1)^1/2, that is to the power one half?


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sapforhire
guest
Jun-05-02, 09:10 AM (EST)
 
3. "RE: Where do I start?"
In response to message #0
 
   Distributing and combining like terms gives you -3x-3=0.
Another way is to think about the (x+1) and realize that small equation appears in each part of the equation; therefore if x+1=0 then the whole equation equals 0


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murat
guest
Jun-06-02, 09:10 AM (EST)
 
4. "RE: Where do I start?"
In response to message #0
 
   it is very simple actually. if you look carefully you should realise that the answer is -1. because if you replace x by -1 every paranthesis becomes 0(zero) so it'satisfies the equation. but if you need the proof...

first lets open the eq.
(4x+4)/2 -(15x+15)/2 + (5x+5)/2 = 0

multiply everything by 2 and get rid of the paranthesis
4x+4-15x-15+5x+5=0

and just solve the eq.
-6x = 6
x= -1


that is it...
hope it helps.

murat


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emily
guest
Jun-07-02, 03:17 PM (EST)
 
5. "RE: Where do I start?"
In response to message #0
 
   i dont know... im not 2 good @ math, even though i have an a in the class! hehe... email me if u REALLY NEEDhelp... im good w/ fractions.. if that helps.


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