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

1 Answer
Feb 8, 2017

Answer:

#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.