Alg3.6.5

Solve $n^{-2}+3n^{-1}+2=0\,$

Let $w=n^{-1}\,$ so that $w^2=n^{-2}\,$ . Now the problem is

$w^2+3w+2=0\,$

$(w+2)(w+1)=0\,$

$w=-2\,$ and $w=-1\,$

Now change back to the original variables,

$n^{-1}=-2\,$ and $n^{-1}=-1\,$

and solve for $n\,$ .

$\frac{1}{n}=-2\,$ and $\frac{1}{n}=-1\,$

Multiply both sides by n

$1=2n\,$ and $1=-n\,$

The solution is

$n=\frac{-1}{2}\,$ and $n=-1\,$