We have to solve the inequation: x^2 - 4x - 12 > 0
First let's solve the quadratic equation x^2 - 4x - 12 = 0
=> x^2 - 6x +2x -12 =0
=> x(x-6) + 2(x -6) =0
=> (x+2)(x-6) =0
Now for x^2 - 4x - 12 > 0
either (x+2) and (x-6) are positive
=> x > -2 and x > 6
=> x > 6
or (x+2) and (x-6) are negative
=> x < -2 and x < 6
=> x < -2
Therefore for x^2 - 4x - 12 > 0 , x should be with less than -2 or greater than 6.
To solve x^2-4x-12 >0.
Since the right side is zero, we express the left as product of two factors. And then see that each factor has opposite signs to have the product negative.
x^2-4x-12 = x^2-6x+2x -12 = x(x-6) +2(x-6) = (x-6)(x+2).
So the x^2-4x-12 < 0 iff (x-6)(x+2) < 0 which can happen only if x takes any value within the interval of the real roots -2 to 6.
Or x^2-4x-12 < 0 for values of x for which -2 < x < 6
First of all, find the values of x (there are 2 because of the grade of inequation) for the inequation is annulled. For this reason, we transform the above inequation into an equation.
x^2 -4x -12 = 0
After that, for finding the roots of the equation, we are using the following formula:
X1=[-b + (b^2 - 4ac)^1/2]/2a, where a=1, b=-4, c =-12
a,b,c being the coefficients of the equation above.
X2= [-b - (b^2 - 4ac)^1/2]/2a
After that, following the rule which says that between the two roots, the values of x have the opposite sign of the "a" coefficient, and outside the roots, the values of X have the same sign with the coefficient "a", we could find the conclusion that inequation is positive on the following intervals
(-infinite, -2) U (6, + infinite)