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Find the number of integers n that satisfy 64 <= n^2 <= 100 + 20n.

 Jun 19, 2024
 #1
avatar+1075 
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Let's split this into two inequalities and solve both to find what x is. 

We have

\(64 \leq n^2\)

and 

\(n^2 \leq 100+20n\)

 

Let's solve the first equation. Square rooting both sides, we get

\(n \geq 8\) or \(n \leq -8\)

 

Solving the second equation, we get that

\(n^2-20n-100 \leq 0\\ 10\left(1-\sqrt{2}\right) \leq n\leq10\left(1+\sqrt{2}\right)\)

 

Merging overlapping intervals, we have

\(8\le \:n\le \:10\left(1+\sqrt{2}\right)\)

 

This rounds to 

\(8\le \:n\le \:24.14\)

 

So our answer is 17. 

Thanks! :)

 Jun 19, 2024
edited by NotThatSmart  Jun 19, 2024

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