How do you solve #1/n^2+1/n=1/(2n^2)# and check for extraneous solutions?

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Answer 1 · 2016-08-17T18:13:35

#n = -1/2#

#1/n^2+1/n=1/(2n^2)+1/(2n^2)+1/n=1/(2n^2)#

so

#1/(2n^2)+cancel(1/(2n^2))+1/n=cancel(1/(2n^2))+0#

so

#1/n(1/(2n)+1)=0#

Finally

#1/(2n)+1=0->n = -1/2#