What is 2x^3-18x<x^2-9 ?

The solution to 2x^3-18x<x^2-9 is x<-3\lor 1/2 <x<3

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2x^3-18x<x^2-9

2 x^{3} - 18 x < x^{2} - 9

x < - 3 or \frac{1}{2} < x < 3

Interval Notation

(- \infty, - 3) \cup (\frac{1}{2}, 3)

Decimal Notation

x < - 3 or 0.5 < x < 3

Solution steps

2 x^{3} - 18 x < x^{2} - 9

Rewrite in standard form

2 x^{3} - 18 x + 9 - x^{2} < 0

Factor

2 x^{3} - 18 x + 9 - x^{2} : (2 x - 1) (x + 3) (x - 3)

(2 x - 1) (x + 3) (x - 3) < 0

Identify the intervals

x < - 3 or \frac{1}{2} < x < 3

\frac{\frac{- 2 x}{x - y}}{\frac{y}{x ^{2}}}

125

\frac{x ^{\frac{1}{2}}}{x ^{\frac{3}{4}}}

\frac{5 x - 1}{3} + 2 \leq \frac{x + 4}{7}

5 x - 9 = - 3 x + 7