# How do you solve (3y^2-9y^3)/(12y^2+5y-2)>=0?

Feb 3, 2018

$\frac{1}{3} \ge y > \frac{1}{4} \mathmr{and} y \text{<} - \frac{2}{3} \mathmr{and} y = 0$

#### Explanation:

If we divide a negative number by another negative number, we get a positive number. If we divide a positive by a positive number, we get also a positive answer. Therefore:

$3 {y}^{2} - 9 {y}^{3} > 0 \mathmr{and} 12 {y}^{2} + 5 y - 2 > 0$

$\mathmr{and}$

$3 {y}^{2} - 9 {y}^{3} < 0 \mathmr{and} 12 {y}^{2} + 5 y - 2 < 0$

$3 {y}^{2} - 9 {y}^{3} > 0$
y^2(3-9y)>0|:y^2 (${y}^{2}$ will always be positive)
3-9y>0|+9y |:9

$\frac{1}{3} > y \mathmr{if} 3 {y}^{2} - 9 {y}^{3} > 0$

Therefore

$\frac{1}{3} \text{<} y \mathmr{if} 3 {y}^{2} - 9 {y}^{3} < 0$

12y^2+5y-2>0|:12
${y}^{2} + \frac{5}{12} y - \frac{1}{6} > 0$
${\left(y + \frac{5}{24}\right)}^{2} - \frac{25}{24} ^ 2 - \frac{1}{6} > 0 | + \frac{25}{24} ^ 2 + \frac{1}{6}$
${\left(y + \frac{5}{24}\right)}^{2} > \frac{121}{576}$
Take the root
$y + \frac{5}{24} > \pm \frac{11}{24} | - \frac{5}{24}$

${y}_{1} > \frac{1}{4} \mathmr{and} {y}_{2} < - \frac{2}{3} \mathmr{if} 12 {y}^{2} + 5 y - 2 > 0$

Therefore:

$\frac{1}{4} \text{>"y_2">} - \frac{2}{3} \mathmr{if} 12 {y}^{2} + 5 y - 2 < 0$

All in all:
$\frac{1}{3} > y > \frac{1}{4} \mathmr{and} y \text{<} - \frac{2}{3}$

Also $y = 0 \mathmr{and} y = \frac{1}{3}$ are answers because of
$3 {y}^{2} - 9 {y}^{3} = 0$

Feb 3, 2018

See below

#### Explanation:

$\frac{3 {y}^{2} - 9 {y}^{3}}{12 {y}^{2} + 5 y - 2} \ge 0$

Multiplying by $12 {y}^{2} + 5 y - 2$ on both sides,
$3 {y}^{2} - 9 {y}^{3} \ge 0$

Dividing by $3 {y}^{2}$ on both sides,
$1 - 3 y \ge 0$

That can be simplified as:
$- 3 y \ge - 1$

To eliminate the negative sign, we have to change the equality sign. Thus, we get it as:
y<=1/3 ; which is your answer.

Feb 3, 2018

The solutions are $y \in \left(- \infty , - \frac{2}{3}\right) \cup \left\{0\right\} \cup \left(\frac{1}{4} , \frac{1}{3}\right]$

#### Explanation:

Solve this inequality with a sign chart

The roots of the denominator

$12 {y}^{2} + 5 y - 2 = 0$ are

${y}_{1} = \frac{- 5 + \sqrt{{5}^{2} - 4 \cdot 12 \cdot \left(- 2\right)}}{2 \cdot 12} = \frac{- 5 + 11}{24} = \frac{6}{24} = \frac{1}{4}$

${y}_{2} = \frac{- 5 - \sqrt{{5}^{2} - 4 \cdot 12 \cdot \left(- 2\right)}}{2 \cdot 12} = \frac{- 5 - 11}{24} = - \frac{16}{24}$
$= - \frac{2}{3}$

The roots of the numerator

$3 {y}^{2} - 9 {y}^{3} = \left(3 {y}^{2}\right) \left(1 - 3 y\right)$

${y}_{3} = 0$

and

${y}_{4} = \frac{1}{3}$

Let $f \left(y\right) = \frac{3 {y}^{2} - 9 {y}^{3}}{12 {y}^{2} + 5 y - 2}$

Construct the sign chart

$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- \frac{2}{3}$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$\frac{1}{4}$$\textcolor{w h i t e}{a a a a}$$\frac{1}{3}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$y + \frac{2}{3}$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$${y}^{2}$$\textcolor{w h i t e}{a a a a a a a}$$+$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$y - \frac{1}{4}$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a}$$-$$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$1 - 3 y$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a}$$+$$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a}$$-$

$\textcolor{w h i t e}{a a a a}$$f \left(y\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a}$$-$$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a}$$-$

Therefore,

$f \left(y\right) \ge 0$ when $y \in \left(- \infty , - \frac{2}{3}\right) \cup \left\{0\right\} \cup \left(\frac{1}{4} , \frac{1}{3}\right]$

graph{(3x^2-9x^3)/(12x^2+5x-2) [-8.02, 6.03, -2.8, 4.23]}