# How do you solve the inequality x^2-x-12>0?

Feb 8, 2017

${x}^{2} - x - 12 > 0 \text{ "=>" "x<"-} 3 \cup x > 4$.

#### Explanation:

Since the right-hand side (RHS) is already 0, we start by factoring the left-hand side (LHS):

$\text{ } {x}^{2} - x - 12 > 0$

$\implies \left(x - 4\right) \left(x + 3\right) > 0$

In its factored form, this inequality tells us that the product of two numbers $\left(x - 4\right)$ and $\left(x + 3\right)$ is positive (greater than $0$).

In order for a product of two terms to be positive, either both terms must be positive or both terms must be negative. So, we require either

$x - 4 > 0 \text{ "nn" } x + 3 > 0$

or

$x - 4 < 0 \text{ "nn" } x + 3 < 0$.

The former simplifies to

$x > 4 \text{ "nn" "x>"-} 3$,

which is only true when $x > 4$. The latter simplifies to

$x < 4 \text{ "nn" "x<"-} 3$,

which is only true when $x < \text{-} 3$. Since either of these situations makes the inequality true, we combine these statements with the logical "or" ($\cup$) to get

${x}^{2} - x - 12 > 0 \text{ "=>" "x<"-} 3 \cup x > 4$.

## Bonus:

We could also use a sign chart to solve this inequality. Once we have the LHS factored, we create our sign chart as follows:

$\underline{\text{ "x" | -3 4 }}$
$\text{ "x-4" |}$
$\underline{\text{ "x+3" | }}$
$\left(x - 4\right) \left(x + 3\right) \text{ } |$

The -3 is there because it's the $x$-value that makes $x + 3$ equal to 0, thus all $x$-values to the left of -3 will make $x + 3$ negative, and all $x$-values to the right will make it positive. (A similar argument follows for the 4.)

Then, fill the two middle rows with $+$ or $-$ signs, depending on where each factor is positive or negative (i.e. for $x$-values in which ranges):

$\underline{\text{ "x" | -3 4 }}$
$\text{ "x-4" | "-" "-" } +$
$\underline{\text{ "x+3" | "-" "+" "+" }}$
$\left(x - 4\right) \left(x + 3\right) \text{ } |$

The final row gets filled by multiplying the signs of all the rows above it:

$\underline{\text{ "x" | -3 4 }}$
$\text{ "x-4" | "-" "-" } +$
$\underline{\text{ "x+3" | "-" "+" "+" }}$
$\left(x - 4\right) \left(x + 3\right) \text{ "|" "+" "-" } +$

This final row tells us that the product $\left(x - 4\right) \left(x + 3\right)$ is positive when $x < \text{-} 3$ or when $x > 4$, which gives us the solution

$\left\{x | x < \text{-} 3 , x > 4\right\}$

which matches the solution from earlier.