9/x-x>= 0

Solution

\frac{9}{x} - x \geq 0

x \leq - 3 or 0 < x \leq 3

Interval Notation

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

Solution steps

\frac{9}{x} - x \geq 0

Rewrite in standard form

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

Factor

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

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

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

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

Identify the intervals

x \leq - 3 or 0 < x \leq 3

2 a^{2} + a - 66

\frac{24 a ^{2} b ^{2} + 12 a b ^{2} + 6 a ^{2} b}{2 ab}

9 - 4 x \leq 3 x - 5

x - d + (x - d)^{2}