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Prove that this is true: x*sqrt(x+1)  sqrt(x^2+1)*(x1) = x^2  2x 1
 one year ago
 one year ago
Prove that this is true: x*sqrt(x+1)  sqrt(x^2+1)*(x1) = x^2  2x 1
 one year ago
 one year ago

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baldymcgee6Best ResponseYou've already chosen the best response.0
\[x*\sqrt(x+1)  \sqrt(x^2+1)*(x1) = x^2  2x 1\]
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
\[x*\sqrt{x+1}  \sqrt{x^2+1}*(x1) = x^2  2x 1\]
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
that's what i thought too...
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
if you plug in x = 1 I get sqrt(2) = 2.
 one year ago

calculusfunctionsBest 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.
 one year ago

calculusfunctionsBest 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?
 one year ago

Jemurray3Best 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.
 one year ago

calculusfunctionsBest 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.
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
@calculusfunctions, no it is not under the root.
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.0
Are you sure you copied the question correctly?
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.0
Could you please double check because I'd really like to help you if I could.
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
yep, don't worry about it. It is very possible that they are not equal.
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
In fact it is so possible that it is true that they are not equal. :)
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
but thanks for your eagerness to help!
 one year ago

calculusfunctionsBest 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.
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.0
No worries. Welcome!
 one year ago
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