sasogeek if i add (x to one) and subtract one from (x plus root of one squared), i get a number twice as big as (x minus itself), what is x? 9 months ago 9 months ago

1. AravindG

First form the equation using the info given

2. sasogeek

3. AravindG

lol why not you try first :)

4. sasogeek

alright... this is what i get :P lol uhmm... add x to one, so, x+1 and subtract one from x plus root of one squared, (x+1)+(x+1^2-1) that's equal to a number twice as big as x minus itself, so (x+1)+(x+1^2-1)=2*(x-x) correct?

5. ParthKohli

6. AravindG

yes

7. sasogeek

nope! LOOOL, look at this line :P and subtract one from x plus root of one squared, (x+1)+(x+1^2-1) where's the root of one squared? it's only one squared, there's no root :P

8. AravindG

$\sqrt{1^2}=1$

9. AravindG

No difference :)

10. ParthKohli

The question is kinda confusing. What does it exactly say? What are you doing with $$x + 1$$ and $$(x + \sqrt{1})^2 - 1$$?

11. sasogeek

that makes all the difference in the initial equation. i think it matters though result may be the same, lol

12. sasogeek

@ParthKohli the full equation is what i have written down, (x+1)+(x+1^2-1)=2*(x-x) but then this part is supposed to be $$\large (x+ \sqrt{1^2}-1)$$ :)

13. ParthKohli

Is the question this?$(x + 1) + (x + \sqrt{1})^2 - 1 = 2(x - x)$If so, you have $$(x + 1) + (x + \sqrt{1})^2 - 1 = 0$$

14. sasogeek

tbh though i shouldn't have let that one out lol, it was supposed to be left for you to figure out :P

15. AravindG

actually When you frame the equation it should be important $(x+1)+(x+\sqrt{1}^2-1)=2 * (x-x)$

16. ParthKohli

Hmm, but I don't think that it'd be $$\sqrt{1}^2$$, but if you insist so...

17. sasogeek

root of 1 squared has only 1 meaning lol, $$\sqrt{1^2}$$

18. ParthKohli

No... this can't be

19. ParthKohli

You will have two solutions then, I guess.

20. sasogeek

nope, just one

21. AravindG

just one

22. AravindG

$\sqrt{1^2}=1$

23. AravindG

Okay so why not solve it straight away @sasogeek ?

24. ParthKohli

Since $$\sqrt{1^2} = \sqrt{1} = \pm 1$$ you may have to solve both equations$(x + 1) +(x - 1)-1 = 0$$(x+1) + (x + 1) -1 =0$

25. AravindG

omg $\sqrt{1} \neq \pm 1$

26. ParthKohli

@AravindG Why not?

27. sasogeek

$$\large \sqrt{1^2}=1^{2*\frac{1}{2}}=1$$

28. AravindG

Because square root is a function having range as positive real numbers

29. ParthKohli

$(-1)^2 = 1 \Rightarrow -1 = \sqrt{1}$$(1)^2 = 1 \Rightarrow 1 = \sqrt{1}$$\therefore \sqrt{1} = \pm 1$@AravindG The domain is positive real number, not the range

30. ParthKohli

Or actually, the domain is nonnegative real numbers.

31. dumbsearch2

Wow.

32. AravindG

@ParthKohli you are having a serious misconception on square roots .

33. sasogeek

you don't have to spam

34. ParthKohli

@AravindG Wasn't my demonstration enough?

35. sasogeek

can we just solve my problem and take this to another thread, lol

36. Peter14

37. AravindG

I can explain it to you separately not here

38. sasogeek

39. AravindG

Lets finish this problem first

40. Peter14

41. sasogeek

lol why thank you :)

42. AravindG

ok @sasogeek proceed :)

43. ParthKohli

If you want to neglect one root of $$1$$, you may. Just consider $$1$$ as the root of 1 if you want to. But seriously, this equation has two possible solutions.

44. dumbsearch2

LOL @Peter14

45. AravindG

x=$$-\frac{1}{2}$$