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  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,272 Thanks: 942 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle n^2 + n = f(n)$ 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 
$n^2n = 2 \implies (n2)(n+1) = 0 \implies n = 2 \in \mathbb{N} \implies n^2+n = 6$ ... or am I missing something? 
September 12th, 2019, 10:24 AM  #4 
Senior Member Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 
I agree. n(n1) = 2 > n = 2, 2^2+2 = 6.


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