Solve sqrt(x ^ 2 5x^4) < x^2

Question

Solve the inequality

x \in (- \frac{5}{5}, 0) \cup (0, \frac{5}{5})

Evaluate

x^{2} \times 5 x^{4} < x^{2}

Multiply

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5 x^{6} < x^{2}

Raise both sides of the inequality to the power of 2

5 x^{6} < (x^{2})^{2}

Evaluate the power

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5 x^{6} < x^{4}

Move the expression to the left side

5 x^{6} - x^{4} < 0

Factor the expression

x^{4} (5 x^{2} - 1) < 0

Separate the inequality into 2 possible cases

{x^{4} > 0 5 x^{2} - 1 < 0 {x^{4} < 0 5 x^{2} - 1 > 0

Solve the inequality

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{x \neq = 0 5 x^{2} - 1 < 0 {x^{4} < 0 5 x^{2} - 1 > 0

Solve the inequality

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{x \neq = 0 - \frac{5}{5} < x < \frac{5}{5} {x^{4} < 0 5 x^{2} - 1 > 0

Since the left-hand side is always positive or 0,and the right-hand side is always 0,the statement is false for any value of x

{x \neq = 0 - \frac{5}{5} < x < \frac{5}{5} {x \in / R 5 x^{2} - 1 > 0

Solve the inequality

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{x \neq = 0 - \frac{5}{5} < x < \frac{5}{5} {x \in / R x \in (- \infty, - \frac{5}{5}) \cup (\frac{5}{5}, + \infty)

Find the intersection

x \in (- \frac{5}{5}, 0) \cup (0, \frac{5}{5}) {x \in / R x \in (- \infty, - \frac{5}{5}) \cup (\frac{5}{5}, + \infty)

Find the intersection

x \in (- \frac{5}{5}, 0) \cup (0, \frac{5}{5}) x \in / R

Solution

x \in (- \frac{5}{5}, 0) \cup (0, \frac{5}{5})

Show Solution