Prove that this is true: x*sqrt(x+1) - sqrt(x^2+1)*(x-1) = x^2 - 2x -1

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.

- baldymcgee6

Prove that this is true: x*sqrt(x+1) - sqrt(x^2+1)*(x-1) = x^2 - 2x -1

- jamiebookeater

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

- baldymcgee6

\[x*\sqrt(x+1) - \sqrt(x^2+1)*(x-1) = x^2 - 2x -1\]

- anonymous

It isn't.

- baldymcgee6

\[x*\sqrt{x+1} - \sqrt{x^2+1}*(x-1) = x^2 - 2x -1\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- baldymcgee6

that's what i thought too...

- anonymous

if you plug in x = 1 I get sqrt(2) = -2.

- baldymcgee6

k thanks.

- calculusfunctions

By proving that L.S. = R.S. we're proving that it is an identity. Which implies that the equation is valid for all x except x > -1. Therefore Jemurray is correct.

- calculusfunctions

Let me ask you though. Do you mean for the (x − 1) to be inside the square root in the second term of the left side of the equation?

- anonymous

Even if he did, it doesn't matter. The fact that the first term on the left is only valid on a restricted domain and the others aren't is enough to show that it can't be an equality.

- calculusfunctions

I know that but I'm grasping for straws trying to understand what he's talking about or even if asked the question correctly.

- baldymcgee6

@calculusfunctions, no it is not under the root.

- calculusfunctions

Are you sure you copied the question correctly?

- calculusfunctions

Could you please double check because I'd really like to help you if I could.

- baldymcgee6

yep, don't worry about it. It is very possible that they are not equal.

- baldymcgee6

In fact it is so possible that it is true that they are not equal. :)

- baldymcgee6

but thanks for your eagerness to help!

- calculusfunctions

You're right! If LS ≠ RS for some values of x, then the equation is not an identity.

- calculusfunctions

No worries. Welcome!

Looking for something else?

Not the answer you are looking for? Search for more explanations.