How do you solve $\frac{16 - x^{2}}{x^{2} - 9} \geq 0$ $(16-x^2)/(x^2-9)>=0$ ?

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How do you solve $\frac{16 - x^{2}}{x^{2} - 9} \geq 0$ $(16-x^2)/(x^2-9)>=0$ ?

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Answer 1 · 2015-06-15T17:35:54

$3 < | x | \leq 4$ $3<|x|<=4$ alternatively $3 < x \leq 4$ $3<x<=4$ or $- 4 \leq x < - 3$ $-4<=x<-3$

Explanation:

For the inequality to be two either $16 - x^{2} \geq 0 and x^{2} - 9 \geq 0$ $16-x^2>=0 and x^2-9>=0$ or $16 - x^{2} \leq 0 and x^{2} - 9 \leq 0$ $16-x^2<=0 and x^2-9<=0$
Note that $| x | = 3$ $|x|=3$ is not defined
Now $16 - x^{2} \geq 0$ $16-x^2>=0$ if $| x | \leq 4$ $|x|<=4$ and $x^{2} - 9 \geq 0$ $x^2-9>=0$ if $| x | > 3$ $|x|>3$ so $3 < | x | \leq 4$ $3<|x|<=4$
Now $16 - x^{2} \leq 0$ $16-x^2<=0$ if $| x | \geq 4$ $|x|>=4$ and $x^{2} - 9 \leq 0$ $x^2-9<=0$ if $| x | < 3$ $|x|<3$ which has no solution

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How do you solve $\frac{16 - x^{2}}{x^{2} - 9} \geq 0$ $(16-x^2)/(x^2-9)>=0$ ?

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