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?

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 x-1=0 is x=1. But, similar to your argument, we can deduce that if x-1=0, then . Thus putting back into the first equation, , implying either x =1 or x=-2.

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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.

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

thx a lot