# How do you solve x^4<4x^2?

Jun 30, 2016

$x \in \left(- 2 , 0\right) \cup \left(0 , 2\right)$

i.e. $- 2 < x < 0$ or $0 < x < 2$

#### Explanation:

First note that if $x = 0$ then ${x}^{4} = 0 = 4 {x}^{2}$, so the inequality is false.

If $x \ne 0$ then ${x}^{2} > 0$ and we can divide both sides of the inequality by ${x}^{2}$ to get:

${x}^{2} < 4$

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Case $\boldsymbol{x > 0}$

Since $y = \sqrt{x}$ is a strictly monotonically increasing function for $x > 0$, we can take the square root of both sides of our simplified inequality to find:

$x < 2$

Hence we have solutions:

$0 < x < 2$

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Case $\boldsymbol{x < 0}$

Note that ${x}^{2} = {\left(- x\right)}^{2}$ and $- x > 0$, so we find:

$- x < 2$

Multiplying both sides by $- 1$ and reversing the inequality, we get:

$x > - 2$

Hence we have solutions:

$- 2 < x < 0$

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Background

The truth or falsity of an inequality is unaltered by any of the following operations:

• Add or subtract the same value from both sides.
• Multiply or divide both sides by the same positive value.
• Multiply or divide both sides by the same negative value and reverse the inequality ($<$ becomes $>$, $\ge$ becomes $\le$, etc.).

More generally:

• Apply the same strictly monotonically increasing function to both sides of the inequality.
• Apply the same strictly monotonically decreasing function to both sides of the inequality and reverse the inequality.