Question

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

Answer 1

`x^2-x-12>0" "=>" "x<"-"3 uu x>4`.

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 `(x-4)` and `(x+3)` 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" "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`. The latter simplifies to

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

which is only true when `x<"-"3`. Since either of these situations makes the inequality true, we combine these statements with the logical "or" (`uu`) to get

`x^2-x-12>0" "=>" "x<"-"3 uu 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:

`ul("            "x"             |            -3               4                   ")`
`"        "x-4"          |"`
`ul("        "x+3"          |                                                   ")`
`(x-4)(x+3)"  "|`

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):

`ul("            "x"             |            -3              4                   ")`
`"        "x-4"          |    "-"          "-"           "+`
`ul("        "x+3"          |    "-"          "+"           "+"          ")`
`(x-4)(x+3)"  "|`

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

`ul("            "x"             |            -3              4                   ")`
`"        "x-4"          |    "-"          "-"           "+`
`ul("        "x+3"          |    "-"          "+"           "+"          ")`
`(x-4)(x+3)" "|"  "+"          "-"            "+`

This final row tells us that the product `(x-4)(x+3)` is positive when `x<"-"3` or when `x>4`, which gives us the solution

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

which matches the solution from earlier.