Solve 8/x^2 >= 16

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

x \in [- \frac{2}{2}, 0) \cup (0, \frac{2}{2}]

Evaluate

\frac{8}{x ^{2}} \geq 16

Find the domain

More Steps

\frac{8}{x ^{2}} \geq 16, x \neq = 0

Move the expression to the left side

\frac{8}{x ^{2}} - 16 \geq 0

Subtract the terms

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\frac{8 - 16 x ^{2}}{x ^{2}} \geq 0

Set the numerator and denominator of \frac{8 - 16 x ^{2}}{x ^{2}} equal to 0 to find the values of x where sign changes may occur

8 - 16 x^{2} = 0 x^{2} = 0

Calculate

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x = \frac{2}{2} x = - \frac{2}{2} x^{2} = 0

The only way a power can be 0 is when the base equals 0

x = \frac{2}{2} x = - \frac{2}{2} x = 0

Determine the test intervals using the critical values

x < - \frac{2}{2} - \frac{2}{2} < x < 0 0 < x < \frac{2}{2} x > \frac{2}{2}

Choose a value form each interval

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

To determine if x < - \frac{2}{2} is the solution to the inequality,test if the chosen value x = - 2 satisfies the initial inequality

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

To determine if - \frac{2}{2} < x < 0 is the solution to the inequality,test if the chosen value x = - \frac{2}{4} satisfies the initial inequality

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

To determine if 0 < x < \frac{2}{2} is the solution to the inequality,test if the chosen value x = \frac{2}{4} satisfies the initial inequality

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

To determine if x > \frac{2}{2} is the solution to the inequality,test if the chosen value x = 2 satisfies the initial inequality

More Steps

x < - \frac{2}{2} is not a solution - \frac{2}{2} < x < 0 is the solution 0 < x < \frac{2}{2} is the solution x > \frac{2}{2} is not a solution

The original inequality is a nonstrict inequality,so include the critical value in the solution

- \frac{2}{2} \leq x < 0 is the solution 0 < x \leq \frac{2}{2} is the solution

The final solution of the original inequality is x \in [- \frac{2}{2}, 0) \cup (0, \frac{2}{2}]

x \in [- \frac{2}{2}, 0) \cup (0, \frac{2}{2}]

Check if the solution is in the defined range

x \in [- \frac{2}{2}, 0) \cup (0, \frac{2}{2}], x \neq = 0

Solution

x \in [- \frac{2}{2}, 0) \cup (0, \frac{2}{2}]

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