How do you solve (16-x^2)/(x^2-9)>=016−x2x2−9≥0?

1 Answer
Phillip E.
Jun 15, 2015

3<|x|<=43<|x|≤4 alternatively 3<x<=43<x≤4 or -4<=x<-3−4≤x<−3

Explanation:

For the inequality to be two either 16-x^2>=0 and x^2-9>=016−x2≥0andx2−9≥0 or 16-x^2<=0 and x^2-9<=016−x2≤0andx2−9≤0
Note that |x|=3|x|=3 is not defined
Now 16-x^2>=016−x2≥0 if |x|<=4|x|≤4 and x^2-9>=0x2−9≥0 if |x|>3|x|>3 so 3<|x|<=43<|x|≤4
Now 16-x^2<=016−x2≤0 if |x|>=4|x|≥4 and x^2-9<=0x2−9≤0 if |x|<3|x|<3 which has no solution

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