## sasogeek Group Title 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? one year ago one year ago

1. AravindG Group Title

First form the equation using the info given

2. sasogeek Group Title

3. AravindG Group Title

lol why not you try first :)

4. sasogeek Group Title

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 Group Title

6. AravindG Group Title

yes

7. sasogeek Group Title

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 Group Title

$\sqrt{1^2}=1$

9. AravindG Group Title

No difference :)

10. ParthKohli Group Title

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 Group Title

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

12. sasogeek Group Title

@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 Group Title

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 Group Title

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 Group Title

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

16. ParthKohli Group Title

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

17. sasogeek Group Title

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

18. ParthKohli Group Title

No... this can't be

19. ParthKohli Group Title

You will have two solutions then, I guess.

20. sasogeek Group Title

nope, just one

21. AravindG Group Title

just one

22. AravindG Group Title

$\sqrt{1^2}=1$

23. AravindG Group Title

Okay so why not solve it straight away @sasogeek ?

24. ParthKohli Group Title

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 Group Title

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

26. ParthKohli Group Title

@AravindG Why not?

27. sasogeek Group Title

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

28. AravindG Group Title

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

29. ParthKohli Group Title

$(-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 Group Title

Or actually, the domain is nonnegative real numbers.

31. dumbsearch2 Group Title

Wow.

32. AravindG Group Title

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

33. sasogeek Group Title

you don't have to spam

34. ParthKohli Group Title

@AravindG Wasn't my demonstration enough?

35. sasogeek Group Title

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

36. Peter14 Group Title

37. AravindG Group Title

I can explain it to you separately not here

38. sasogeek Group Title

39. AravindG Group Title

Lets finish this problem first

40. Peter14 Group Title

41. sasogeek Group Title

lol why thank you :)

42. AravindG Group Title

ok @sasogeek proceed :)

43. ParthKohli Group Title

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 Group Title

LOL @Peter14

45. AravindG Group Title

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