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What is the greatest integer $n$ such that $n^2 - 11n +24 \leq 17n - 8$?

 Feb 19, 2024
 #1
avatar+1622 
+1

What is the greatest integer n such that \(n^2 - 11n +24 \leq 17n - 8\)?

Combining like terms: \(n^2-28n+32\leq0\)

Using quadratic formula n = \({28\pm4\sqrt{41}\over2}=14\pm2\sqrt{41}\). Since our quadratic must be nonnegative, when factored (x - r)(x - s) where r and s are our roots, they must both be positive or both negative, so we take the extreme intervals: \((-\inf,14-2\sqrt{41}]\) U \([14 + 2\sqrt{41}, \inf)\)

 Feb 19, 2024
 #2
avatar+128406 
+1

Borrowing from proyaop's answer

 

14 + 2sqrt (41)  ≈ 26.8

This is  the larger root  and all  integer x's > than this will make the polynomial > 0

So.....the greatest integer that makes  this le to 0   is 26

 

cool cool cool

 Feb 19, 2024
edited by CPhill  Feb 19, 2024

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