# How can I find the values of x that satisfy x^2 - 4x -12 > 0

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

x^2 - 4x - 12 > 0

To solve the inequality, we will use the same rules of solvingthe equalities.

Let us factor:

( x-6) ( x+ 2) > 0

Then, we have two cases:

case(1):

(x-6) > 0 and ( x+ 2) >0

Then, x > 6  and x > -2

Then, x = ( 6, inf)

case(2):

(x-6) < 0    and  ( x+2) < 0

==> x < 6   and    x < -2

Then x = (-inf, -2)

Then the solution is:

x = ( -inf, -2) U ( 6, inf)

OR x = R - [ -2, 6]

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We need to solve x^2 - 4x -12 > 0.

Now x^2 - 4x -12 > 0

=> x^2 – 6x + 2x – 12 >0

=> x (x – 6) + 2 (x – 6)>0

=> (x – 6) (x + 2) >0

Now this is true either when both (x – 6) and (x + 2) are greater than zero or when both (x – 6) and (x + 2) are less than zero.

For both (x – 6) and (x + 2) to be greater than 0 x > 6.

For both (x – 6) and (x + 2) to be less than 0 x < -2

Therefore the values of x less than -2 and greater than 6 satisfy the inequation.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find the solutions for the equation x^2-4x+12 > 0.

We first factorise the left side  expression x^2-4x-12.

x^2-4x+12 = x^2-6x+2x-12

x^2-4x+12 = x(x-6) +2(x-6)

x^2-4x+12 = (x-6) (x+2).

Therefore  x^2-4x-12> 0 implies (x+2)(x-6) > 0.

Therefore both factors should be of the same sign.

Therefore x+2 < 0 and x-6 < 0 ,

Or (x+2)> 0 and (x-6) > 0.

This is possinble only when  x < -2   Or  x > 6 , the x^2-4x-12 > 0.

Or x must be in the interval (-infinity -2) Or in the interval (6  infinity).