How do you solve #1/(n^2 + 11n + 30) = 1/(n+5) -4 #?

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Answer 1 · 2017-10-16T12:41:28

No real solution

#1/(n^2+11n+30)=1/(n+5)#

#:.1/((n+5)(n+6))=1/(n+5)#

multiply both sides by #(n+5)(n+6)#

#:.1=n+6-4(n+5)(n+6)#

#;.n+6-4(n+5)(n+6)=1#

#;.n+6-4n^2-44n-120=1#

#:.-4n^2-43n-120=1#

#:.-4n^2-43n-119=0#

multiply both sides by #-1#

#;.4n^2+43n+119=0#

#:.n=(-b+-sqrt(b^2-4ac))/(2a)#

#;.n=(-(43)+-sqrt((43^2)-4(4)(119)))/(2(4))#

#:.-43+-sqrt(1849-1904)/8#

#:.-43+-sqrt(-55)/8#

No real solution.