How do you solve #\frac { 5} { n } - \frac { 6} { n ^ { 3} - 2n ^ { 2} } = \frac { n ^ { 2} + 5n - 1} { n ^ { 3} - 2n ^ { 2} }#?

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Answer 1 · 2017-06-26T13:18:27

#n=15/8+-sqrt305/8#

#5/n-6/(n^3-2n^2)=(n^2+5n-1)/(n^3-2n^2)#

#hArr(5n(n-2))/(nxxn(n-2))-6/(n^2(n-2))=(n^2+5n-1)/(n^2(n-2))#

Observe that #n# cannot take values #0# and #2#.

Now multiplying each term by #n^2(n-2)#, we get

#5n^2-10n-6=n^2+5n-1#

or #4n^2-15n-5=0#

and using quadratic formula

#n=(-(-15)+-sqrt((-15)^2-4xx4xx(-5)))/(4xx2)#

= #(15+-sqrt(225+80))/8#

= #15/8+-sqrt305/8#