Solve (x^2 6x-7)/|x^4| <0

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

x \in (- \infty, 0) \cup (0, \frac{3 252}{6})

Evaluate

\frac{x ^{2} \times 6 x - 7}{∣ x ^{4} ∣} < 0

Find the domain

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\frac{x ^{2} \times 6 x - 7}{∣ x ^{4} ∣} < 0, x \neq = 0

Simplify

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\frac{6 x ^{3} - 7}{x ^{4}} < 0

Set the numerator and denominator of \frac{6 x ^{3} - 7}{x ^{4}} equal to 0 to find the values of x where sign changes may occur

6 x^{3} - 7 = 0 x^{4} = 0

Calculate

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x = \frac{3 252}{6} x^{4} = 0

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

x = \frac{3 252}{6} x = 0

Determine the test intervals using the critical values

x < 0 0 < x < \frac{3 252}{6} x > \frac{3 252}{6}

Choose a value form each interval

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

To determine if x < 0 is the solution to the inequality,test if the chosen value x = - 1 satisfies the initial inequality

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x < 0 is the solution x_{2} = 1 x_{3} = 2

To determine if 0 < x < \frac{3 252}{6} is the solution to the inequality,test if the chosen value x = 1 satisfies the initial inequality

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x < 0 is the solution 0 < x < \frac{3 252}{6} is the solution x_{3} = 2

To determine if x > \frac{3 252}{6} is the solution to the inequality,test if the chosen value x = 2 satisfies the initial inequality

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x < 0 is the solution 0 < x < \frac{3 252}{6} is the solution x > \frac{3 252}{6} is not a solution

The original inequality is a strict inequality,so does not include the critical value ,the final solution is x \in (- \infty, 0) \cup (0, \frac{3 252}{6})

x \in (- \infty, 0) \cup (0, \frac{3 252}{6})

Check if the solution is in the defined range

x \in (- \infty, 0) \cup (0, \frac{3 252}{6}), x \neq = 0

Solution

x \in (- \infty, 0) \cup (0, \frac{3 252}{6})

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