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

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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]**

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.**

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).