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?
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AravindG
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First form the equation using the info given
sasogeek
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go right ahead :)
AravindG
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lol why not you try first :)
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?
ParthKohli
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(Hey, congrads!)
AravindG
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yes
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
AravindG
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\[\sqrt{1^2}=1\]
AravindG
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No difference :)
ParthKohli
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The question is kinda confusing. What does it exactly say? What are you doing with \(x + 1\) and \((x + \sqrt{1})^2 - 1\)?
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
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 \(\large (x+ \sqrt{1^2}-1) \)
:)
ParthKohli
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Is the question this?\[(x + 1) + (x + \sqrt{1})^2 - 1 = 2(x - x)\]If so, you have \((x + 1) + (x + \sqrt{1})^2 - 1 = 0\)
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
AravindG
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actually
When you frame the equation it should be important
\[(x+1)+(x+\sqrt{1}^2-1)=2 * (x-x)\]
ParthKohli
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Hmm, but I don't think that it'd be \(\sqrt{1}^2\), but if you insist so...
sasogeek
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root of 1 squared has only 1 meaning lol, \(\sqrt{1^2} \)
ParthKohli
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No... this can't be
ParthKohli
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You will have two solutions then, I guess.
sasogeek
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nope, just one
AravindG
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just one
AravindG
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\[\sqrt{1^2}=1\]
AravindG
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Okay so why not solve it straight away @sasogeek ?
ParthKohli
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Since \(\sqrt{1^2} = \sqrt{1} = \pm 1\) you may have to solve both equations\[(x + 1) +(x - 1)-1 = 0\]\[(x+1) + (x + 1) -1 =0\]
AravindG
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omg \[\sqrt{1} \neq \pm 1\]
ParthKohli
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@AravindG Why not?
sasogeek
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\(\large \sqrt{1^2}=1^{2*\frac{1}{2}}=1 \)
AravindG
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Because square root is a function having range as positive real numbers
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
ParthKohli
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Or actually, the domain is nonnegative real numbers.
dumbsearch2
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Wow.
AravindG
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@ParthKohli you are having a serious misconception on square roots .
sasogeek
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you don't have to spam
ParthKohli
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@AravindG Wasn't my demonstration enough?
sasogeek
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can we just solve my problem and take this to another thread, lol
Peter14
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so where was this problem from?
AravindG
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I can explain it to you separately not here
sasogeek
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my head, @Peter14
AravindG
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Lets finish this problem first
Peter14
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you have an interesting head.
sasogeek
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lol why thank you :)
AravindG
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ok @sasogeek proceed :)
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.
dumbsearch2
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LOL @Peter14
AravindG
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@ParthKohli I can make a separate thread for your query .
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
sasogeek
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x=\(-\frac{1}{2} \)