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baldymcgee6
 3 years ago
Prove that this is true: x*sqrt(x+1)  sqrt(x^2+1)*(x1) = x^2  2x 1
baldymcgee6
 3 years ago
Prove that this is true: x*sqrt(x+1)  sqrt(x^2+1)*(x1) = x^2  2x 1

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baldymcgee6
 3 years ago
Best ResponseYou've already chosen the best response.0\[x*\sqrt(x+1)  \sqrt(x^2+1)*(x1) = x^2  2x 1\]

baldymcgee6
 3 years ago
Best ResponseYou've already chosen the best response.0\[x*\sqrt{x+1}  \sqrt{x^2+1}*(x1) = x^2  2x 1\]

baldymcgee6
 3 years ago
Best ResponseYou've already chosen the best response.0that's what i thought too...

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1if you plug in x = 1 I get sqrt(2) = 2.

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.0By 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
 3 years ago
Best ResponseYou've already chosen the best response.0Let 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?

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1Even 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
 3 years ago
Best ResponseYou've already chosen the best response.0I know that but I'm grasping for straws trying to understand what he's talking about or even if asked the question correctly.

baldymcgee6
 3 years ago
Best ResponseYou've already chosen the best response.0@calculusfunctions, no it is not under the root.

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.0Are you sure you copied the question correctly?

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.0Could you please double check because I'd really like to help you if I could.

baldymcgee6
 3 years ago
Best ResponseYou've already chosen the best response.0yep, don't worry about it. It is very possible that they are not equal.

baldymcgee6
 3 years ago
Best ResponseYou've already chosen the best response.0In fact it is so possible that it is true that they are not equal. :)

baldymcgee6
 3 years ago
Best ResponseYou've already chosen the best response.0but thanks for your eagerness to help!

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.0You're right! If LS ≠ RS for some values of x, then the equation is not an identity.

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.0No worries. Welcome!
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