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September 12th, 2019, 05:30 AM   #1
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Evaluate expression

if $\displaystyle n^2 -n =2 \:$ , find $\displaystyle n^2 +n\;$, $\displaystyle n\in \mathbb{N}.$
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September 12th, 2019, 09:21 AM   #2
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Originally Posted by idontknow View Post
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
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September 12th, 2019, 09:26 AM   #3
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$n^2-n = 2 \implies (n-2)(n+1) = 0 \implies n = 2 \in \mathbb{N} \implies n^2+n = 6$

... or am I missing something?
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September 12th, 2019, 10:24 AM   #4
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I agree. n(n-1) = 2 -> n = 2, 2^2+2 = 6.
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