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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?

Mathematics
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First form the equation using the info given
go right ahead :)
lol why not you try first :)

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Other answers:

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?
(Hey, congrads!)
yes
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
\[\sqrt{1^2}=1\]
No difference :)
The question is kinda confusing. What does it exactly say? What are you doing with \(x + 1\) and \((x + \sqrt{1})^2 - 1\)?
that makes all the difference in the initial equation. i think it matters though result may be the same, lol
@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 \(\large (x+ \sqrt{1^2}-1) \) :)
Is the question this?\[(x + 1) + (x + \sqrt{1})^2 - 1 = 2(x - x)\]If so, you have \((x + 1) + (x + \sqrt{1})^2 - 1 = 0\)
tbh though i shouldn't have let that one out lol, it was supposed to be left for you to figure out :P
actually When you frame the equation it should be important \[(x+1)+(x+\sqrt{1}^2-1)=2 * (x-x)\]
Hmm, but I don't think that it'd be \(\sqrt{1}^2\), but if you insist so...
root of 1 squared has only 1 meaning lol, \(\sqrt{1^2} \)
No... this can't be
You will have two solutions then, I guess.
nope, just one
just one
\[\sqrt{1^2}=1\]
Okay so why not solve it straight away @sasogeek ?
Since \(\sqrt{1^2} = \sqrt{1} = \pm 1\) you may have to solve both equations\[(x + 1) +(x - 1)-1 = 0\]\[(x+1) + (x + 1) -1 =0\]
omg \[\sqrt{1} \neq \pm 1\]
@AravindG Why not?
\(\large \sqrt{1^2}=1^{2*\frac{1}{2}}=1 \)
Because square root is a function having range as positive real numbers
\[(-1)^2 = 1 \Rightarrow -1 = \sqrt{1}\]\[(1)^2 = 1 \Rightarrow 1 = \sqrt{1}\]\[\therefore \sqrt{1} = \pm 1\]@AravindG The domain is positive real number, not the range
Or actually, the domain is nonnegative real numbers.
Wow.
@ParthKohli you are having a serious misconception on square roots .
you don't have to spam
@AravindG Wasn't my demonstration enough?
can we just solve my problem and take this to another thread, lol
so where was this problem from?
I can explain it to you separately not here
my head, @Peter14
Lets finish this problem first
you have an interesting head.
lol why thank you :)
ok @sasogeek proceed :)
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.
@ParthKohli I can make a separate thread for your query .
it's not that hard u guys -_-, it boils down to 2x+1=0, hard? i don't think so xD
x=\(-\frac{1}{2} \)

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