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What is an easy and fast way to solve the problem without going through all these possibilities: a) $n^2-9n+20>0, 16-n^2>0$, b) $n^2-9n+20>0, 16-n^2<0$, c) $n^2-9n+20<0, 16-n^2>0$, and d) $n^2-9n+20<0, 16-n^2<0$? I got the correct answer, but it was very time consuming.

Find the product of all real numbers $/n^2-9n+20/=/16-n^2/$.

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  • $\begingroup$ Are these slashes meant to indicate absolute value? $\endgroup$ – mvw May 21 '16 at 20:13
  • $\begingroup$ It isn't clear what it meant by "all real numbers in an equation". Are you asking about roots of an equation? But you wrote several inequalities. Then you have an equation involving absolute values? $\endgroup$ – hardmath May 21 '16 at 20:16
  • $\begingroup$ I get $(1/2) \cdot 4 = 2$. $\endgroup$ – mvw May 21 '16 at 20:17
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    $\begingroup$ Both sides have a common factor $n-4$. Dividing that out gives $|n-5|=|n+4|$, so the two solutions are $n=4,\frac{1}{2}$. $\endgroup$ – almagest May 21 '16 at 20:17
  • $\begingroup$ @mvw You are right. It is 2. My question is how we can quickly eliminate conditions b), c), and d) without actually trying them. $\endgroup$ – user321527 May 21 '16 at 20:20

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