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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?
 10 months ago
 10 months 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?
 10 months ago
 10 months ago

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AravindGBest ResponseYou've already chosen the best response.3
First form the equation using the info given
 10 months ago

AravindGBest ResponseYou've already chosen the best response.3
lol why not you try first :)
 10 months ago

sasogeekBest ResponseYou've already chosen the best response.0
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+1^21) that's equal to a number twice as big as x minus itself, so (x+1)+(x+1^21)=2*(xx) correct?
 10 months ago

sasogeekBest ResponseYou've already chosen the best response.0
nope! LOOOL, look at this line :P and subtract one from x plus root of one squared, (x+1)+(x+1^21) where's the root of one squared? it's only one squared, there's no root :P
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
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

sasogeekBest ResponseYou've already chosen the best response.0
that makes all the difference in the initial equation. i think it matters though result may be the same, lol
 10 months ago

sasogeekBest ResponseYou've already chosen the best response.0
@ParthKohli the full equation is what i have written down, (x+1)+(x+1^21)=2*(xx) but then this part is supposed to be \(\large (x+ \sqrt{1^2}1) \) :)
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
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

sasogeekBest ResponseYou've already chosen the best response.0
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

AravindGBest ResponseYou've already chosen the best response.3
actually When you frame the equation it should be important \[(x+1)+(x+\sqrt{1}^21)=2 * (xx)\]
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
Hmm, but I don't think that it'd be \(\sqrt{1}^2\), but if you insist so...
 10 months ago

sasogeekBest ResponseYou've already chosen the best response.0
root of 1 squared has only 1 meaning lol, \(\sqrt{1^2} \)
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
No... this can't be
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
You will have two solutions then, I guess.
 10 months ago

AravindGBest ResponseYou've already chosen the best response.3
Okay so why not solve it straight away @sasogeek ?
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
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

AravindGBest ResponseYou've already chosen the best response.3
omg \[\sqrt{1} \neq \pm 1\]
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
@AravindG Why not?
 10 months ago

sasogeekBest ResponseYou've already chosen the best response.0
\(\large \sqrt{1^2}=1^{2*\frac{1}{2}}=1 \)
 10 months ago

AravindGBest ResponseYou've already chosen the best response.3
Because square root is a function having range as positive real numbers
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
\[(1)^2 = 1 \Rightarrow 1 = \sqrt{1}\]\[(1)^2 = 1 \Rightarrow 1 = \sqrt{1}\]\[\therefore \sqrt{1} = \pm 1\]@AravindG The domain is positive real number, not the range
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
Or actually, the domain is nonnegative real numbers.
 10 months ago

AravindGBest ResponseYou've already chosen the best response.3
@ParthKohli you are having a serious misconception on square roots .
 10 months ago

sasogeekBest ResponseYou've already chosen the best response.0
you don't have to spam
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
@AravindG Wasn't my demonstration enough?
 10 months ago

sasogeekBest ResponseYou've already chosen the best response.0
can we just solve my problem and take this to another thread, lol
 10 months ago

Peter14Best ResponseYou've already chosen the best response.0
so where was this problem from?
 10 months ago

AravindGBest ResponseYou've already chosen the best response.3
I can explain it to you separately not here
 10 months ago

AravindGBest ResponseYou've already chosen the best response.3
Lets finish this problem first
 10 months ago

Peter14Best ResponseYou've already chosen the best response.0
you have an interesting head.
 10 months ago

sasogeekBest ResponseYou've already chosen the best response.0
lol why thank you :)
 10 months ago

AravindGBest ResponseYou've already chosen the best response.3
ok @sasogeek proceed :)
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
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

AravindGBest ResponseYou've already chosen the best response.3
@ParthKohli I can make a separate thread for your query .
 10 months ago

sasogeekBest ResponseYou've already chosen the best response.0
it's not that hard u guys _, it boils down to 2x+1=0, hard? i don't think so xD
 10 months ago
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