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 September 12th, 2019, 05:30 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91 Evaluate expression if $\displaystyle n^2 -n =2 \:$ , find $\displaystyle n^2 +n\;$, $\displaystyle n\in \mathbb{N}.$
September 12th, 2019, 09:21 AM   #2
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Quote:
 Originally Posted by idontknow if $\displaystyle n^2 -n =2 \:$ , find $\displaystyle n^2 +n\;$, $\displaystyle n\in \mathbb{N}.$
$\displaystyle n^2 - n = 2$
$\displaystyle n^2 + n = f(n)$
where f(n) is, so far, an undetermined function.

Subtract these:
$\displaystyle -2n = 2 - f(n)$

$\displaystyle f(n) = 2n + 2$

So...
$\displaystyle n^2 + n = 2n + 2$

Is this what you were looking for? (And yes, there is probably a better way to do this.)

-Dan

 September 12th, 2019, 09:26 AM #3 Math Team     Joined: Jul 2011 From: Texas Posts: 3,016 Thanks: 1600 $n^2-n = 2 \implies (n-2)(n+1) = 0 \implies n = 2 \in \mathbb{N} \implies n^2+n = 6$ ... or am I missing something? Thanks from topsquark, idontknow and DarnItJimImAnEngineer
 September 12th, 2019, 10:24 AM #4 Senior Member   Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 I agree. n(n-1) = 2 -> n = 2, 2^2+2 = 6. Thanks from topsquark and idontknow

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