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Find all values of x such that (x^2 - 11x + 30)/(2x^2 - x - 6) >= 0.

 Feb 5, 2022
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\(\frac{ (x^2 - 11x + 30)}{(2x^2 - x - 6) }\ge 0. \)

 

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

 Feb 5, 2022

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