+118690
\(\frac{ (x^2 - 11x + 30)}{(2x^2 - x - 6) }\ge 0. \)
It wil be bigger than zero when the top and bottom are both positive, or both negartive.
f(x)=x^2-11x+30, is a concave up parabola
f(x)=(x-5)(x-6)
the roots are +5 and +6.
It will be negative between these values and positive outside
Now look at the bottom - another concave up parabola
g(x)=2x^2 -x -6
i need two numbers that multiply to -12 and add to -1. That would be -4 and +3 -x will be replaced with -4x+3x
g(x)=2x^2 -4x+3x -6
g(x)=2x(x-2)+3(x-2)
g(x)=(2x+3)(x-2)
roots are -1.5, and 2 Since this is the denominator of a fraction, x cannot actually be -1.5 or 2. (you cannot divide by 0)
I suggest you draw this on a rough number line. You should be able to see that they are never both negaitve at the same time.
However they are both positve when
\(x<-1.5,\qquad -2
All thses x vaues make the original statement valid.
Here is a pic
