How do you solve #(5n - 1) (n+1) = 0#?

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Answer 1 · 2016-12-23T17:18:07

#n=1/5# #or n=-1#

#(5n-1)(n+1)=0#

#5n^2+4n-1=0#

#(5n-1)=0#

#5n=1#

#n=1/5#

or

#(n+1)=0#

#n=-1#

substitute # n=1/5#

#5(1/5)^2+4(1/5)-1=0#

#5(1/25)+4/5-1=0#

#5/25+4/5-1=0#

#1/5+4/5-1=0#

#5/5-1=0#

#1-1=0#

substitute # n=-1#

#5n^2+4n-1=0#

#5(-1)^2+(4(-1)-1=0#

#5+(-4)-1=0#

#1-1=0#

Answer 2 · 2016-12-25T19:36:08

#n = 1/5" or "n=-1#

This quadratic is already in factor form.

#(5n-1)(n+1)=0#

If the product of 2 factors is 0, one of them must be 0.

Consider each factor equal to 0.

If #5n-1=0color(white)(..................) "if " n+1=0#

#rarr" "5n = 1color(white)(..............)rarr" "n=-1#

#rarr" "n = 1/5#