What is ((n^2+2))/((n^2-1))+2<0 ?

The solution to ((n^2+2))/((n^2-1))+2<0 is -1<n<0\lor 0<n<1

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((n^2+2))/((n^2-1))+2<0

\frac{( n ^{2} + 2 )}{( n ^{2} - 1 )} + 2 < 0

- 1 < n < 0 or 0 < n < 1

Interval Notation

(- 1, 0) \cup (0, 1)

Solution steps

\frac{n ^{2} + 2}{n ^{2} - 1} + 2 < 0

Rewrite in standard form

\frac{n ^{2}}{n ^{2} - 1} < 0

Factor

\frac{n ^{2}}{n ^{2} - 1} : \frac{n ^{2}}{( n + 1 ) ( n - 1 )}

\frac{n ^{2}}{( n + 1 ) ( n - 1 )} < 0

Identify the intervals

- 1 < n < 0 or 0 < n < 1

3 - x^{2} < 0

\frac{( x ^{2} \cdot ( x - 2 ) ( 5 - x ) )}{( ( x ^{2} - 4 ) ( x - 1 ) )} > 0

\frac{( 6 x - x ^{2} )}{( ( x + 9 ) ^{2} )} \leq 0

\frac{2}{3 x} > - 6 x

\frac{( x + 1 ) ( x - 1 ) ( x + 2 )}{( x ^{2} + 3 x + 2 ) ( x + 5 )} \leq 0

What is ((n^2+2))/((n^2-1))+2<0 ?
The solution to ((n^2+2))/((n^2-1))+2<0 is -1<n<0\lor 0<n<1