## sasogeek 2 years ago 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?

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

46. sasogeek

it's not that hard u guys -_-, it boils down to 2x+1=0, hard? i don't think so xD

47. sasogeek

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