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dont need weird mathematics :) \[f(x)=\sqrt{x^2-2x+10}-\sqrt{x^2-2x+2}\] find all \(x \in \mathbb{R}\) such that \(f(x) \in \mathbb{N}\)

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x=1,\[\frac{ -1-\sqrt{5} }{2 }\], \[\frac{-1+\sqrt{5} }{2 }\]
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Other answers:

let me check it :)
u are close :) but dont tell the answer plz...so others can think on that :)
Ok fine. Thanks :)
So only when x=1 or are there any other value..... I think yes because irrational-irrational can be integer
some other values...
I mean are there any other values of x
yeah there are.
dont post ur solution :) just values of x
Ok
So I delete it and send u msg for the vaues of x
no give the values of x here :)
|dw:1347712962196:dw|
oh crap! I misread x^2 as x^3 :\
So was my answer correct @nipunmalhotra93
@mukushla can u PLZ CHECK MY ANSWER
completely right :)
so i think nipun got it also :)
Actually f(x)=2 and 1
yup
GREAT QUESTION
Took me a lot to do this......
:D the font is too small I thought it was x^3 lol :D
everyone is one here...lol
@mukushla CAN U PLZ GIVE SOME QUESTION LIKE THIS PLZ.......
of course
GREAT
@sauravshakya I'd like to know your approach. Would you mind telling it? This is how I got it. \[\sqrt{p+10}-\sqrt{p+2}=A\] A is natural so, \[p+10=A ^{2}+p+2+2A \sqrt{p+2}\] \[10-A^2-2=2A \sqrt{p+2}\] so, \[\frac{ 4}{A }-\frac{ A }{ 2 }=\sqrt{p+2}\] So, A can only be 1 or 2 (as A is natural) for \[\sqrt{p+2}\] to be real. Put p=x^2-2x and solving \[\sqrt{p+2}=1 or \sqrt{p+2}=7/2\] , we get the answer. This is a good question that's why I wanna know your method. :)
Rationalize and u will get|dw:1347715001113:dw|
Right?
\[f(x)=\sqrt{x^2-2x+10}-\sqrt{x^2-2x+2}=\frac{8}{\sqrt{x^2-2x+10}+\sqrt{x^2-2x+2}}\]\[f(x)=\frac{8}{\sqrt{(x-1)^2+9}+\sqrt{(x-1)^2+1}}\le\frac{8}{\sqrt{9}+\sqrt{1}}=2\]so f(x)=1 or 2 same as saura's method
That will make it easy to solve
opps. Now, when f(x)=1 (x^2-2x+10)^1/2 - (x^2-2x+2)^1/2=1........i (x^2-2x+10)^1/2 + (x^2-2x+2)^1/2=8........ii
So, 2*(x^2-2x+10)^1/2=9
now, solve for x
thats it

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