What is ((9-x^2))/x >0 ?

The solution to ((9-x^2))/x >0 is x<-3\lor 0<x<3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

((9-x^2))/x >0

\frac{( 9 - x ^{2} )}{x} > 0

x < - 3 or 0 < x < 3

Interval Notation

(- \infty, - 3) \cup (0, 3)

Solution steps

\frac{9 - x ^{2}}{x} > 0

Factor

\frac{9 - x ^{2}}{x} : \frac{- ( x + 3 ) ( x - 3 )}{x}

\frac{- ( x + 3 ) ( x - 3 )}{x} > 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x + 3 ) ( x - 3 ) ) ( - 1 )}{x} < 0 \cdot (- 1)

Simplify

\frac{( x + 3 ) ( x - 3 )}{x} < 0

Identify the intervals

x < - 3 or 0 < x < 3

3 x^{2} - 11 x + 2 < 0

12 - (x^{2} - 101) < - 74 + 6 x^{2}

\frac{( 8 + 4 x )}{( 4 x + x ^{2} )} \leq \frac{2}{x} + \frac{3}{( 4 + x )}

\frac{x ^{2} + ( 4 x ^{2} )}{( ( x - 2 ) ^{2} )} \leq 5

(3 x + 1)^{2} + 4 x < 9 x^{2} + x - 4