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 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?
 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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AravindG
 one year ago
Best ResponseYou've already chosen the best response.3First form the equation using the info given

AravindG
 one year ago
Best ResponseYou've already chosen the best response.3lol why not you try first :)

sasogeek
 one year ago
Best ResponseYou've already chosen the best response.0alright... 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?

sasogeek
 one year ago
Best ResponseYou've already chosen the best response.0nope! 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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0The question is kinda confusing. What does it exactly say? What are you doing with \(x + 1\) and \((x + \sqrt{1})^2  1\)?

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

sasogeek
 one year ago
Best 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 \(\large (x+ \sqrt{1^2}1) \) :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Is 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
 one year ago
Best ResponseYou've already chosen the best response.0tbh though i shouldn't have let that one out lol, it was supposed to be left for you to figure out :P

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

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

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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0No... this can't be

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0You will have two solutions then, I guess.

AravindG
 one year ago
Best ResponseYou've already chosen the best response.3Okay so why not solve it straight away @sasogeek ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Since \(\sqrt{1^2} = \sqrt{1} = \pm 1\) you may have to solve both equations\[(x + 1) +(x  1)1 = 0\]\[(x+1) + (x + 1) 1 =0\]

AravindG
 one year ago
Best ResponseYou've already chosen the best response.3omg \[\sqrt{1} \neq \pm 1\]

sasogeek
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \sqrt{1^2}=1^{2*\frac{1}{2}}=1 \)

AravindG
 one year ago
Best ResponseYou've already chosen the best response.3Because square root is a function having range as positive real numbers

ParthKohli
 one year ago
Best 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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Or actually, the domain is nonnegative real numbers.

AravindG
 one year ago
Best ResponseYou've already chosen the best response.3@ParthKohli you are having a serious misconception on square roots .

sasogeek
 one year ago
Best ResponseYou've already chosen the best response.0you don't have to spam

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0@AravindG Wasn't my demonstration enough?

sasogeek
 one year ago
Best ResponseYou've already chosen the best response.0can we just solve my problem and take this to another thread, lol

Peter14
 one year ago
Best ResponseYou've already chosen the best response.0so where was this problem from?

AravindG
 one year ago
Best ResponseYou've already chosen the best response.3I can explain it to you separately not here

AravindG
 one year ago
Best ResponseYou've already chosen the best response.3Lets finish this problem first

Peter14
 one year ago
Best ResponseYou've already chosen the best response.0you have an interesting head.

AravindG
 one year ago
Best ResponseYou've already chosen the best response.3ok @sasogeek proceed :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0If 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.

AravindG
 one year ago
Best ResponseYou've already chosen the best response.3@ParthKohli I can make a separate thread for your query .

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