At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

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