Solve \(\frac{2x^5}{x^{2}-9}>0\)

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

x \in (- 3, 0) \cup (3, + \infty)

Evaluate

\frac{2 x ^{5}}{x ^{2} - 9} > 0

Find the domain

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\frac{2 x ^{5}}{x ^{2} - 9} > 0, x \in (- \infty, - 3) \cup (- 3, 3) \cup (3, + \infty)

Set the numerator and denominator of \frac{2 x ^{5}}{x ^{2} - 9} equal to 0 to find the values of x where sign changes may occur

2 x^{5} = 0 x^{2} - 9 = 0

Calculate

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x = 0 x^{2} - 9 = 0

Calculate

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x = 0 x = 3 x = - 3

Determine the test intervals using the critical values

x < - 3 - 3 < x < 0 0 < x < 3 x > 3

Choose a value form each interval

x_{1} = - 4 x_{2} = - 2 x_{3} = 2 x_{4} = 4

To determine if x < - 3 is the solution to the inequality,test if the chosen value x = - 4 satisfies the initial inequality

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x < - 3 is not a solution x_{2} = - 2 x_{3} = 2 x_{4} = 4

To determine if - 3 < x < 0 is the solution to the inequality,test if the chosen value x = - 2 satisfies the initial inequality

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x < - 3 is not a solution - 3 < x < 0 is the solution x_{3} = 2 x_{4} = 4

To determine if 0 < x < 3 is the solution to the inequality,test if the chosen value x = 2 satisfies the initial inequality

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x < - 3 is not a solution - 3 < x < 0 is the solution 0 < x < 3 is not a solution x_{4} = 4

To determine if x > 3 is the solution to the inequality,test if the chosen value x = 4 satisfies the initial inequality

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x < - 3 is not a solution - 3 < x < 0 is the solution 0 < x < 3 is not a solution x > 3 is the solution

The original inequality is a strict inequality,so does not include the critical value ,the final solution is x \in (- 3, 0) \cup (3, + \infty)

x \in (- 3, 0) \cup (3, + \infty)

Check if the solution is in the defined range

x \in (- 3, 0) \cup (3, + \infty), x \in (- \infty, - 3) \cup (- 3, 3) \cup (3, + \infty)

Solution

x \in (- 3, 0) \cup (3, + \infty)

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