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

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

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
It isn't.
 2 years ago

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

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
that's what i thought too...
 2 years ago

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

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
k thanks.
 2 years ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.0
I know that but I'm grasping for straws trying to understand what he's talking about or even if asked the question correctly.
 2 years ago

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

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

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

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

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

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

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

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.0
No worries. Welcome!
 2 years ago
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