+1858 Find the number of integers n that satisfy 64 <= n^2 <= 100 + 20n.
+947 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! :)
