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

##### 1 Answer

#### Explanation:

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

#" "x^2-x-12>0#

#=>(x-4)(x+3)>0#

In its factored form, this inequality tells us that the *product* of two numbers

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" "nn" "x+3>0#

or

#x-4<0" "nn" "x+3<0# .

The former simplifies to

#x>4" "nn" "x>"-"3# ,

which is only true when

#x<4" "nn" "x<"-"3# ,

which is only true when *either* of these situations makes the inequality true, we combine these statements with the logical "or" (

## 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:

The -3 is there because it's the *(A similar argument follows for the 4.)*

Then, fill the two middle rows with

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

This final row tells us that the *product*

#{x|x<"-"3,x>4}#

which matches the solution from earlier.