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

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- AravindG

First form the equation using the info given

- sasogeek

go right ahead :)

- AravindG

lol why not you try first :)

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## More answers

- sasogeek

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

(Hey, congrads!)

- AravindG

yes

- sasogeek

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

\[\sqrt{1^2}=1\]

- AravindG

No difference :)

- ParthKohli

The question is kinda confusing. What does it exactly say? What are you doing with \(x + 1\) and \((x + \sqrt{1})^2 - 1\)?

- sasogeek

that makes all the difference in the initial equation. i think it matters though result may be the same, lol

- sasogeek

@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

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

tbh though i shouldn't have let that one out lol, it was supposed to be left for you to figure out :P

- AravindG

actually
When you frame the equation it should be important
\[(x+1)+(x+\sqrt{1}^2-1)=2 * (x-x)\]

- ParthKohli

Hmm, but I don't think that it'd be \(\sqrt{1}^2\), but if you insist so...

- sasogeek

root of 1 squared has only 1 meaning lol, \(\sqrt{1^2} \)

- ParthKohli

No... this can't be

- ParthKohli

You will have two solutions then, I guess.

- sasogeek

nope, just one

- AravindG

just one

- AravindG

\[\sqrt{1^2}=1\]

- AravindG

Okay so why not solve it straight away @sasogeek ?

- ParthKohli

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

omg \[\sqrt{1} \neq \pm 1\]

- ParthKohli

@AravindG Why not?

- sasogeek

\(\large \sqrt{1^2}=1^{2*\frac{1}{2}}=1 \)

- AravindG

Because square root is a function having range as positive real numbers

- ParthKohli

\[(-1)^2 = 1 \Rightarrow -1 = \sqrt{1}\]\[(1)^2 = 1 \Rightarrow 1 = \sqrt{1}\]\[\therefore \sqrt{1} = \pm 1\]@AravindG The domain is positive real number, not the range

- ParthKohli

Or actually, the domain is nonnegative real numbers.

- dumbsearch2

Wow.

- AravindG

@ParthKohli you are having a serious misconception on square roots .

- sasogeek

you don't have to spam

- ParthKohli

@AravindG Wasn't my demonstration enough?

- sasogeek

can we just solve my problem and take this to another thread, lol

- anonymous

so where was this problem from?

- AravindG

I can explain it to you separately not here

- sasogeek

my head, @Peter14

- AravindG

Lets finish this problem first

- anonymous

you have an interesting head.

- sasogeek

lol why thank you :)

- AravindG

ok @sasogeek proceed :)

- ParthKohli

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

LOL @Peter14

- AravindG

@ParthKohli I can make a separate thread for your query .

- sasogeek

it's not that hard u guys -_-, it boils down to 2x+1=0, hard? i don't think so xD

- sasogeek

x=\(-\frac{1}{2} \)

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