# Solve the inequality 30/x-1 < x+2?

$x \setminus \in \left(\setminus \frac{- 1 - \setminus \sqrt{129}}{2} , 1\right) \setminus \cup \left(\setminus \frac{- 1 + \setminus \sqrt{129}}{2} , \setminus \infty\right)$

#### Explanation:

$\setminus \frac{30}{x - 1} < x + 2$

$\setminus \frac{30}{x - 1} - \left(x + 2\right) < 0$

$\setminus \frac{30 - \left(x + 2\right) \left(x - 1\right)}{x - 1} < 0$

$\setminus \frac{30 - {x}^{2} - x + 2}{x - 1} < 0$

$\setminus \frac{- {x}^{2} - x + 32}{x - 1} < 0$

$\setminus \frac{{x}^{2} + x - 32}{x - 1} > 0$

Using quadratic formula to find the roots of ${x}^{2} + x - 32 = 0$ as follows

$x = \setminus \frac{- 1 \setminus \pm \setminus \sqrt{{1}^{2} - 4 \left(1\right) \left(- 32\right)}}{2 \left(1\right)}$

$x = \setminus \frac{- 1 \setminus \pm \setminus \sqrt{129}}{2}$

$\setminus \therefore \setminus \frac{\left(x + \setminus \frac{1 + \setminus \sqrt{129}}{2}\right) \left(x + \setminus \frac{1 - \setminus \sqrt{129}}{2}\right)}{x - 1} > 0$

Solving above inequality, we get

$x \setminus \in \left(\setminus \frac{- 1 - \setminus \sqrt{129}}{2} , 1\right) \setminus \cup \left(\setminus \frac{- 1 + \setminus \sqrt{129}}{2} , \setminus \infty\right)$

Jun 26, 2018

color(blue)((-1/2-1/2sqrt(129),1)uuu(-1/2+1/2sqrt(129),oo)

#### Explanation:

$\frac{30}{x - 1} < x + 2$

subtract $\left(x + 2\right)$ from both sides:

$\frac{30}{x - 1} - x - 2 < 0$

Simplify $L H S$

$\frac{- {x}^{2} - x + 32}{x - 1} < 0$

Find roots of numerator:

$- {x}^{2} - x + 32 = 0$

$x = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(- 1\right) \left(32\right)}}{2 \left(- 1\right)}$

$x = \frac{1 \pm \sqrt{129}}{-} 2$

$x = - \frac{1}{2} + \frac{1}{2} \sqrt{129}$

$x = - \frac{1}{2} - \frac{1}{2} \sqrt{129}$

For $x > - \frac{1}{2} + \frac{1}{2} \sqrt{129}$

$- {x}^{2} - x + 32 < 0$

For $x < - \frac{1}{2} + \frac{1}{2} \sqrt{129}$

$- {x}^{2} - x + 32 > 0$

For $x > - \frac{1}{2} - \frac{1}{2} \sqrt{129}$

$- {x}^{2} - x + 32 > 0$

For $x < - \frac{1}{2} - \frac{1}{2} \sqrt{129}$

$- {x}^{2} - x + 32 < 0$

Root of $x - 1$

$x - 1 = 0 \implies x = 1$

For: $x > 1$

$x - 1 > 0$

For $x < 1$

$x - 1 < 0$

Check for:

$\frac{+}{-}$, $\frac{-}{+}$

This gives us:

$- \frac{1}{2} - \frac{1}{2} \sqrt{129} < x < 1$

$- \frac{1}{2} + \frac{1}{2} \sqrt{129} < x < \infty$

In interval notation this is:

$\left(- \frac{1}{2} - \frac{1}{2} \sqrt{129} , 1\right) \bigcup \left(- \frac{1}{2} + \frac{1}{2} \sqrt{129} , \infty\right)$