Question

Solve equation, check for extraneous solutions

Answer 1

Start with

$1=\dfrac{1}{n^2}+\dfrac{n^2-4n-5}{3n^2}$

We observe that both fractions are not defined if $n=0$. So, we will assume $n \neq 0$.

We multiply both numerator and denominator of the first fraction by 3 and we sum the two fractions:

$1=\dfrac{3}{3n^2}+\dfrac{n^2-4n-5}{3n^2} = \dfrac{n^2-4n-2}{3n^2}$

We multiply both sides by $3n^2$:

$n^2-4n-2=3n^2$

We move everything to one side and solve the quadratic equation:

$2n^2+4n+2=0 \iff 2(n^2+2n+1)=0 \iff 2(n+1)^2=0 \iff n+1=0 \iff n=-1$

We check the solution:

$1=\dfrac{1}{1}+\dfrac{1+4-5}{3} \iff 1 = 1+0$

which is true