# Solve the inequation x^2 - 4x - 12 > 0

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X^2-4x-12>0

First we will solve the inequality just like we solve any quadric equation.

x^2-4x-12>0

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

That means that both terms are either negative or both positive: OR

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

x>6 **and** x>-2

==> the x belongs to the interval (6, inf)

OR

(x-6)<0 **and** (x+2)<0

x<6 **and** x<-2

Then x belongs to the interval (-inf, -2)

From both solution we observe that x belongs to R except for the interval [-2,6]

then x = R-[-2,6]

First of all, we'll solve the qudratic equation:

x^2 -4x -12 = 0

After that, to find out the roots of the equation, we'll use the quadratic formula:

x1 = [-b + sqrt(b^2 - 4ac)]/2a, where a=1, b=-4, c =-12

x2 = [-b + sqrt(b^2 - 4ac)]/2a,

after calculation

x1=6, x2=-2

After that, following the rule which says that between the two roots, the values of function have the opposite sign of the "a" coefficient, and outside the roots, the values of the function have the same sign with the coefficient "a", we could find the conclusion that inequation is positive on the following intervals:

x belongs to the interval (-infinite, -2) U (6, + infinite).

The in-equation x^2 - 4x - 12 > 0 has to be solved.

x^2 - 4x - 12 > 0

x^2 - 6x + 2x - 12 > 0

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

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

This is true when either both x + 2 and x - 6 are greater than 0 or when both x + 2 and x - 6 are less than 0

x + 2 > 0 and x - 6 > 0

x > -2 and x > 6

This is true when x > 6

x + 2 < 0 and x - 6 < 0

x < -2 and x < 6

This is true when x < -2

The solution of the given inequation is `(-oo, -2)U(6, oo)`

To solvw x^2-4x-12 = 0.

Solution:

x^2-4x-12 = 0 could be expressed as (x^2-4x+4) -4 -12 = 0. Or

(x-2)^2 - 16 = 0. Or

(x-2)^2 -4^2 = 0. Or

{(x-2)+4}{(x-2)-4 } = 0. as a^2-b^2 = (a+b)(a-b).

(x+2)(x-6) = 0. Or

x+2 = 0 Or x-6 = 0.

x+2 = 0 gives x=-2. And x-6 = 0 gives, x = 6.