# How do you solve 2X^3-x^4<=0?

Jan 6, 2017

The values for $x$ that are solutions are $x \ge 2 \text{ and } x \le 0$

Written another way $x \in \left(- \infty , 0 \textcolor{w h i t e}{.}\right] \text{ and } x \in \left[\textcolor{w h i t e}{.} 2 , + \infty\right)$

#### Explanation:

Given:$\text{ } 2 {x}^{3} - {x}^{4} \le 0$

${x}^{3} \left(2 - x\right) \le 0$

Suppose $x \ge 2$ then $\left(2 - x\right) \le 0$ so ${x}^{3} \left(2 - x\right) \le 0$
Consequently $x \ge 2$ is part of the answer

Suppose $x = 0$ then ${x}^{3} \left(2 - x\right) = 0$ which satisfies the $\le 0$ criterion. Consequently $x = 0$ is part of the solution.

Suppose $x < 0$ then ${x}^{3} < 0$ and $\left(2 - x\right) > 0$ so their product is $< 0$ thus satisfying the criterion $\le 0$

So the value for $x$ that are solutions are $x \ge 2 \text{ and } x \le 0$