# How do you solve and write the following in interval notation: x^2 < 3x + 18?

Mar 28, 2017

$\left(- 3 , 6\right)$

#### Explanation:

Rearrange inequality to have zero on the right side.

$\Rightarrow {x}^{2} - 3 x - 18 < 0$

Factorise the quadratic on the left side.

$\Rightarrow \left(x - 6\right) \left(x + 3\right) < 0$

For the left side to be negative, that is < 0 , then.

$x - 6 > 0 \textcolor{red}{\text{ and }} x + 3 < 0$

$\Rightarrow x > 6 \textcolor{red}{\text{ and " x<-3larrcolor(blue)" impossible}}$

$\textcolor{m a \ge n t a}{\text{ OR}}$

$x - 6 < 0 \textcolor{red}{\text{ and }} x + 3 > 0$

$\Rightarrow x < 6 \textcolor{red}{\text{ and " x> -3larrcolor(blue)" solution}}$

$\Rightarrow x \in \left(- 3 , 6\right)$

Mar 28, 2017

$x \in \left(- 3 , 6\right)$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} < 3 x + 18$

Re-write as:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 3 x - 18 < 0$

$\textcolor{w h i t e}{\text{XXX}} \left(x - 6\right) \left(x + 3\right) < 0$

Note that $x \ne 0$ since that would make the left side equal to $0$.

The left side is a quadratic dividing the domain of $x$ (after excluding $\left\{- 3 , 6\right\}$ into 3 subsets:

{: (," | ",underline(x < -3),color(white)("X"),underline(x in (-3,6)),color(white)("X"),ul(x > 6)), (ul"Sample "," | ",ul(x=-4),,ul(x=0),,ul(x=7)), (x^2," | ",16,,0,,49), (ul(3x+18)," | ",ul(6),,ul(18),,ul(39)), (x^2 < 3x+18?," | ","False",,"True",,"False") :}