# How do you solve x^3>=9x^2?

##### 1 Answer
Dec 27, 2016

$x \in \left[9 , \infty\right) \cup \left\{0\right\}$

#### Explanation:

${x}^{3} \ge 9 {x}^{2}$

$\implies {x}^{3} - 9 {x}^{2} \ge 0$

$\implies {x}^{2} \left(x - 9\right) \ge 0$

${x}^{2} \left(x - 9\right)$ has two roots: $x = 0$ and $x = 9$. We will check what happens in the intervals on each side of them.

Case 1: $x \in \left(- \infty , 0\right)$

$\implies {x}^{2} > 0 \mathmr{and} x - 9 < 0$

$\implies {x}^{2} \left(x - 9\right) < 0$

Case 2: $x \in \left(0 , 9\right)$

$\implies {x}^{2} > 0 \mathmr{and} x - 9 < 0$

$\implies {x}^{2} \left(x - 9\right) < 0$

Case 3: $x \in \left(9 , \infty\right)$

$\implies {x}^{2} > 0 \mathmr{and} x - 9 > 0$

$\implies {x}^{2} \left(x - 9\right) > 0$

Thus we have ${x}^{2} \left(x - 9\right) \ge 0$ if $x = 0$ or $x \in \left[9 , \infty\right)$