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First form the equation using the info given

go right ahead :)

lol why not you try first :)

(Hey, congrads!)

yes

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

No difference :)

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

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

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

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

Lets finish this problem first

you have an interesting head.

lol why thank you :)

LOL @Peter14

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