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Mathematics 19 Online
OpenStudy (baldymcgee6):

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

OpenStudy (baldymcgee6):

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

OpenStudy (anonymous):

It isn't.

OpenStudy (baldymcgee6):

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

OpenStudy (baldymcgee6):

that's what i thought too...

OpenStudy (anonymous):

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

OpenStudy (baldymcgee6):

k thanks.

OpenStudy (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.

OpenStudy (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?

OpenStudy (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.

OpenStudy (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.

OpenStudy (baldymcgee6):

@calculusfunctions, no it is not under the root.

OpenStudy (calculusfunctions):

Are you sure you copied the question correctly?

OpenStudy (calculusfunctions):

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

OpenStudy (baldymcgee6):

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

OpenStudy (baldymcgee6):

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

OpenStudy (baldymcgee6):

but thanks for your eagerness to help!

OpenStudy (calculusfunctions):

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

OpenStudy (calculusfunctions):

No worries. Welcome!

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