How do you solve #5/(n^3+5n^2)=4/(n+5)+1/n^2# and check for extraneous solutions?

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Answer 1 · 2016-11-04T00:26:35

Let's start by determining the LCD, or the Least Common Denominator.

#5/(n^2(n + 5)) = 4/( n+ 5) + 1/n^2#

#5/(n^2(n + 5)) = (4(n^2))/(n^2(n + 5)) + (n + 5)/(n^2(n + 5))#

We can now eliminate the denominators.

#5 = 4n^2 + n + 5#

#0 = 4n^2 + n#

#0 = n(4n + 1)#

#n = 0 and -1/4#

However, #n = 0# is extraneous since it is one of the restrictions on the original equation.

Hence, #n = -1/4# is the only actual solution.

Hopefully this helps!