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wch2eraq35
Joined: 04 Jan 2005Posts: 13
Location: HK,Tai Po

Posted: Mon Jan 10, 2005 11:17 pm Post subject: Need help…….!



can i solve the equation in that way?
Put this back to the third step…
x=0 or 2 or no real roots.. But 2 doesn’t satisfy the equation…
anyone know what i did wrong or if my method cannot be used that way?





Ellen
Frequent VisitorJoined: 28 Oct 2003Posts: 197
Location: CUHK

Posted: Tue Jan 11, 2005 2:59 pm Post subject: Re: Need help…….!



What you have deduced from above is that if x is real and satisfies then x has got to be 0 or 2. But it does not guarantee that if x= 0 or 2 then x will satisfy the equation.
For example, we know that the root to the equation x1=0 is x=1. But, similar to your argument, we can deduce that if x1=0, then . Thus putting back into the first equation, , implying either x =1 or x=2.
[unparseable or potentially dangerous latex formula]





kazuhiko
Joined: 14 Nov 2004Posts: 3
Location: Salesian English School

Posted: Tue Jan 11, 2005 11:45 pm Post subject:



x=0 or no real roots Therefore, (*) has only one solution is x=0_________________
I think, therefore I am.





wch2eraq35
Joined: 04 Jan 2005Posts: 13
Location: HK,Tai Po

Posted: Wed Jan 12, 2005 12:25 pm Post subject:



thx a lot





