Help with this inequation! x^2 - 4x -12>0
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 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]
Just simply the quadratic equation first which is:
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