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sasogeek
 2 years 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?
sasogeek
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.3First form the equation using the info given

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

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

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

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

ParthKohli
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.3omg \[\sqrt{1} \neq \pm 1\]

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

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

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

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

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

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

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

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

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

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

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

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

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

sasogeek
 2 years 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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