Let and such that . Let be a function such that . Prove that has at least one root

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- March 23rd 2011, 10:40 AMmathfunFunction and root
Let and such that . Let be a function such that . Prove that has at least one root

- March 23rd 2011, 10:57 AMTheChaz
What have you tried so far?

"Prove that..." is*your*responsibility, and while many of us (myself included) are willing and able to help, you have to show some effort besides registering for a forum and copying the question down in LaTex markup. - March 23rd 2011, 11:02 AMmathfun
Ok, seeing the hypothesis i said: there are roots of the equation because its equivalent to . I set d and e in the given relation. and then i get stuck.

- March 25th 2011, 12:38 AMmathfun
Let be such that it is the solution of the equation where .

I set x0 in the given relation

Now one of the roots of the last equation are and we get the wanted.

Is it correct? - March 25th 2011, 07:27 PMTheChaz
- March 26th 2011, 12:32 AMmathfun
Actually it has 2 roots, right? Plaese see the solution i gave in post No. 5

- March 26th 2011, 05:20 AMTheChaz
I don't believe so. If (b - 1)^2 = 4ac, then your "two" solutions are actually the same.

- March 26th 2011, 05:26 AMmathfun
but

- March 26th 2011, 06:08 AMTheChaz
Hah! All this time I thought it was >=!!

ok you win :) - March 26th 2011, 06:10 AMmathfun
- March 26th 2011, 07:34 AMworc3247
Looks fine to me.