For what values of x, the inequality is true? x^2-4x=<12
First let us factorize:
==> x-6=<0 and (x+2)=>0
==> x=<6 and x=>-2
==> x belongs to [-2,6]
OR (x-6)=>0 and (x+2)=<0
==> x=>6 and x=<-2 which is impossible.
Then the solution is x belongs to [-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,
We know that the expression is negative between the roots, because the values of function have the opposite sign of the "a" coefficient, which is positive.
The inequality is negative when x belongs to the interval [-2,6].
x^2-4x<=12 is theinequality. To detrmine x for which the inequatlity holds:
We shift aubtract 12 from both sides , then the equality becoms:
x^2-4x-12 < 0. Or
x^2-6x+2x-12 <0. Or
x(x-6)+2(x-6) < 0. Or
(x-6)(x+2) < 0. For this product of the factors (x-6)(x+2) should be negative. Or x should lie between the roots -2 and 6.