mukushla
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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mukushla
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@sauravshakya @ganeshie8 @eliassaab @TuringTest
nipunmalhotra93
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x=1,\[\frac{ -1-\sqrt{5} }{2 }\],
\[\frac{-1+\sqrt{5} }{2 }\]
nipunmalhotra93
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is that it?
mukushla
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let me check it :)
mukushla
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u are close :) but dont tell the answer plz...so others can think on that :)
nipunmalhotra93
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Ok fine. Thanks :)
sauravshakya
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So only when x=1 or are there any other value..... I think yes because irrational-irrational can be integer
mukushla
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some other values...
sauravshakya
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I mean are there any other values of x
mukushla
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yeah there are.
mukushla
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dont post ur solution :) just values of x
sauravshakya
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Ok
sauravshakya
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So I delete it and send u msg for the vaues of x
mukushla
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no give the values of x here :)
sauravshakya
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|dw:1347712962196:dw|
nipunmalhotra93
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oh crap! I misread x^2 as x^3 :\
sauravshakya
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So was my answer correct @nipunmalhotra93
sauravshakya
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@mukushla can u PLZ CHECK MY ANSWER
mukushla
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completely right :)
mukushla
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so i think nipun got it also :)
sauravshakya
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Actually f(x)=2 and 1
mukushla
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yup
sauravshakya
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GREAT QUESTION
sauravshakya
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Took me a lot to do this......
nipunmalhotra93
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:D the font is too small I thought it was x^3 lol :D
mukushla
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everyone is one here...lol
sauravshakya
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@mukushla CAN U PLZ GIVE SOME QUESTION LIKE THIS PLZ.......
mukushla
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of course
sauravshakya
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GREAT
nipunmalhotra93
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@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. :)
sauravshakya
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Rationalize and u will get|dw:1347715001113:dw|
sauravshakya
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Right?
mukushla
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\[f(x)=\sqrt{x^2-2x+10}-\sqrt{x^2-2x+2}=\frac{8}{\sqrt{x^2-2x+10}+\sqrt{x^2-2x+2}}\]\[f(x)=\frac{8}{\sqrt{(x-1)^2+9}+\sqrt{(x-1)^2+1}}\le\frac{8}{\sqrt{9}+\sqrt{1}}=2\]so f(x)=1 or 2 same as saura's method
sauravshakya
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That will make it easy to solve
sauravshakya
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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
sauravshakya
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So, 2*(x^2-2x+10)^1/2=9
sauravshakya
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now, solve for x
mukushla
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thats it