## anonymous 4 years ago dont need weird mathematics :) $f(x)=\sqrt{x^2-2x+10}-\sqrt{x^2-2x+2}$ find all $$x \in \mathbb{R}$$ such that $$f(x) \in \mathbb{N}$$

1. anonymous

@sauravshakya @ganeshie8 @eliassaab @TuringTest

2. nipunmalhotra93

x=1,$\frac{ -1-\sqrt{5} }{2 }$, $\frac{-1+\sqrt{5} }{2 }$

3. nipunmalhotra93

is that it?

4. anonymous

let me check it :)

5. anonymous

u are close :) but dont tell the answer plz...so others can think on that :)

6. nipunmalhotra93

Ok fine. Thanks :)

7. anonymous

So only when x=1 or are there any other value..... I think yes because irrational-irrational can be integer

8. anonymous

some other values...

9. anonymous

I mean are there any other values of x

10. anonymous

yeah there are.

11. anonymous

dont post ur solution :) just values of x

12. anonymous

Ok

13. anonymous

So I delete it and send u msg for the vaues of x

14. anonymous

no give the values of x here :)

15. anonymous

|dw:1347712962196:dw|

16. nipunmalhotra93

oh crap! I misread x^2 as x^3 :\

17. anonymous

So was my answer correct @nipunmalhotra93

18. anonymous

@mukushla can u PLZ CHECK MY ANSWER

19. anonymous

completely right :)

20. anonymous

so i think nipun got it also :)

21. anonymous

Actually f(x)=2 and 1

22. anonymous

yup

23. anonymous

GREAT QUESTION

24. anonymous

Took me a lot to do this......

25. nipunmalhotra93

:D the font is too small I thought it was x^3 lol :D

26. anonymous

everyone is one here...lol

27. anonymous

@mukushla CAN U PLZ GIVE SOME QUESTION LIKE THIS PLZ.......

28. anonymous

of course

29. anonymous

GREAT

30. nipunmalhotra93

@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. :)

31. anonymous

Rationalize and u will get|dw:1347715001113:dw|

32. anonymous

Right?

33. anonymous

$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

34. anonymous

That will make it easy to solve

35. anonymous

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

36. anonymous

So, 2*(x^2-2x+10)^1/2=9

37. anonymous

now, solve for x

38. anonymous

thats it