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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?

  • 10 months ago
  • 10 months ago

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  1. AravindG
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    First form the equation using the info given

    • 10 months ago
  2. sasogeek
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    go right ahead :)

    • 10 months ago
  3. AravindG
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    lol why not you try first :)

    • 10 months ago
  4. sasogeek
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    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?

    • 10 months ago
  5. ParthKohli
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    (Hey, congrads!)

    • 10 months ago
  6. AravindG
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    yes

    • 10 months ago
  7. sasogeek
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    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

    • 10 months ago
  8. AravindG
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    \[\sqrt{1^2}=1\]

    • 10 months ago
  9. AravindG
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    No difference :)

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

    • 10 months ago
  11. sasogeek
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    that makes all the difference in the initial equation. i think it matters though result may be the same, lol

    • 10 months ago
  12. sasogeek
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    @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 \(\large (x+ \sqrt{1^2}-1) \) :)

    • 10 months ago
  13. ParthKohli
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    Is the question this?\[(x + 1) + (x + \sqrt{1})^2 - 1 = 2(x - x)\]If so, you have \((x + 1) + (x + \sqrt{1})^2 - 1 = 0\)

    • 10 months ago
  14. sasogeek
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    tbh though i shouldn't have let that one out lol, it was supposed to be left for you to figure out :P

    • 10 months ago
  15. AravindG
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    actually When you frame the equation it should be important \[(x+1)+(x+\sqrt{1}^2-1)=2 * (x-x)\]

    • 10 months ago
  16. ParthKohli
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    Hmm, but I don't think that it'd be \(\sqrt{1}^2\), but if you insist so...

    • 10 months ago
  17. sasogeek
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    root of 1 squared has only 1 meaning lol, \(\sqrt{1^2} \)

    • 10 months ago
  18. ParthKohli
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    No... this can't be

    • 10 months ago
  19. ParthKohli
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    You will have two solutions then, I guess.

    • 10 months ago
  20. sasogeek
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    nope, just one

    • 10 months ago
  21. AravindG
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    just one

    • 10 months ago
  22. AravindG
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    \[\sqrt{1^2}=1\]

    • 10 months ago
  23. AravindG
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    Okay so why not solve it straight away @sasogeek ?

    • 10 months ago
  24. ParthKohli
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    Since \(\sqrt{1^2} = \sqrt{1} = \pm 1\) you may have to solve both equations\[(x + 1) +(x - 1)-1 = 0\]\[(x+1) + (x + 1) -1 =0\]

    • 10 months ago
  25. AravindG
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    omg \[\sqrt{1} \neq \pm 1\]

    • 10 months ago
  26. ParthKohli
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    @AravindG Why not?

    • 10 months ago
  27. sasogeek
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    \(\large \sqrt{1^2}=1^{2*\frac{1}{2}}=1 \)

    • 10 months ago
  28. AravindG
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    Because square root is a function having range as positive real numbers

    • 10 months ago
  29. ParthKohli
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    \[(-1)^2 = 1 \Rightarrow -1 = \sqrt{1}\]\[(1)^2 = 1 \Rightarrow 1 = \sqrt{1}\]\[\therefore \sqrt{1} = \pm 1\]@AravindG The domain is positive real number, not the range

    • 10 months ago
  30. ParthKohli
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    Or actually, the domain is nonnegative real numbers.

    • 10 months ago
  31. dumbsearch2
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    Wow.

    • 10 months ago
  32. AravindG
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    @ParthKohli you are having a serious misconception on square roots .

    • 10 months ago
  33. sasogeek
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    you don't have to spam

    • 10 months ago
  34. ParthKohli
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    @AravindG Wasn't my demonstration enough?

    • 10 months ago
  35. sasogeek
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    can we just solve my problem and take this to another thread, lol

    • 10 months ago
  36. Peter14
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    so where was this problem from?

    • 10 months ago
  37. AravindG
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    I can explain it to you separately not here

    • 10 months ago
  38. sasogeek
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    my head, @Peter14

    • 10 months ago
  39. AravindG
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    Lets finish this problem first

    • 10 months ago
  40. Peter14
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    you have an interesting head.

    • 10 months ago
  41. sasogeek
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    lol why thank you :)

    • 10 months ago
  42. AravindG
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    ok @sasogeek proceed :)

    • 10 months ago
  43. ParthKohli
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    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.

    • 10 months ago
  44. dumbsearch2
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    LOL @Peter14

    • 10 months ago
  45. AravindG
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    @ParthKohli I can make a separate thread for your query .

    • 10 months ago
  46. sasogeek
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    it's not that hard u guys -_-, it boils down to 2x+1=0, hard? i don't think so xD

    • 10 months ago
  47. sasogeek
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    x=\(-\frac{1}{2} \)

    • 10 months ago
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