# Help with this inequation! x^2 - 4x -12>0

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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, forf 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 calculation

X1=6, X2=-2

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

(-ininite, -2) U (6, + infinite)

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

x^2 - 4x -12>0

We can factorize the polynomial on the left

x^2 - 4x - 12

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

= x(x - 6) + 2(x - 6)

= (x + 2)(x - 6)

Now (x + 2)(x - 6) is greater than 0 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 gives us x >6 as all numbers greater than 6 are greater than -2

x + 2 < 0 and x - 6 < 0

x < -2 and x < 6

This gives use x < -2 as all numbers lesser than -2 are lesser than 6

The solution of the inequality is all real numbers except those that lie in the set [-2, 6]

The solution is R - [2, -6]

x^2-4x-12>0

Just simply the quadratic equation first which is:

(x+a)(x+b)>0

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

Suppose (x-6)(x+2)=0

x=6 or x=-2

Thus, the quadratic curve cuts the X-axis at x=6 and x=-2, and the curve has a minimum point as coefficient of x^2 is positive

For (x-6)(x+2)>0, i.e. foy y>0, we needed to find the range of values of x for which the curve is on or above the X-axis

Thus, the range of values of x is **x>6 or x<-2**