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sasogeek

  • one year 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?

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  1. AravindG
    • one year ago
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    First form the equation using the info given

  2. sasogeek
    • one year ago
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    go right ahead :)

  3. AravindG
    • one year ago
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    lol why not you try first :)

  4. sasogeek
    • one year ago
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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?

  5. ParthKohli
    • one year ago
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    (Hey, congrads!)

  6. AravindG
    • one year ago
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    yes

  7. sasogeek
    • one year ago
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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

  8. AravindG
    • one year ago
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    \[\sqrt{1^2}=1\]

  9. AravindG
    • one year ago
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    No difference :)

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

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

  12. sasogeek
    • one year ago
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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 \(\large (x+ \sqrt{1^2}-1) \) :)

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

  14. sasogeek
    • one year ago
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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

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

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

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

  18. ParthKohli
    • one year ago
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    No... this can't be

  19. ParthKohli
    • one year ago
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    You will have two solutions then, I guess.

  20. sasogeek
    • one year ago
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    nope, just one

  21. AravindG
    • one year ago
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    just one

  22. AravindG
    • one year ago
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    \[\sqrt{1^2}=1\]

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

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

  25. AravindG
    • one year ago
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    omg \[\sqrt{1} \neq \pm 1\]

  26. ParthKohli
    • one year ago
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    @AravindG Why not?

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

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

  29. ParthKohli
    • one year ago
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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

  30. ParthKohli
    • one year ago
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    Or actually, the domain is nonnegative real numbers.

  31. dumbsearch2
    • one year ago
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    Wow.

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

  33. sasogeek
    • one year ago
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    you don't have to spam

  34. ParthKohli
    • one year ago
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    @AravindG Wasn't my demonstration enough?

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

  36. Peter14
    • one year ago
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    so where was this problem from?

  37. AravindG
    • one year ago
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    I can explain it to you separately not here

  38. sasogeek
    • one year ago
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    my head, @Peter14

  39. AravindG
    • one year ago
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    Lets finish this problem first

  40. Peter14
    • one year ago
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    you have an interesting head.

  41. sasogeek
    • one year ago
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    lol why thank you :)

  42. AravindG
    • one year ago
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    ok @sasogeek proceed :)

  43. ParthKohli
    • one year ago
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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.

  44. dumbsearch2
    • one year ago
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    LOL @Peter14

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

  46. sasogeek
    • one year ago
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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

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

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