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I am trying to understand this problem but my textbook does not offer a good explanation,it asks what the integer part of the square root is.

$\sqrt{n^2+6n}$ , $n \in N$*

ROBOTICS
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  • Hint: find an integer $k$ (in terms of $n$) such that $k^2 \le n^2 + 6n < (k+1)^2$ (this may only work for "large" $n$). Hint2: it is better if you state specifically where you don't understand/do not agree with the presentation of the solution in your book, or we might give the same solutions as your book and fail to help you. – player3236 Oct 22 '20 at 13:08
  • Well it was on how they got to (k+1)^2 because I knew that [x] is defined as k<= x <= k+1 – ROBOTICS Oct 22 '20 at 13:10
  • More precisely $[x]$ is defined as the integer $k$ where $k \le x < k+1$. Taking squares we have $k^2 \le n^2+6n < (k+1)^2$. Now I see your question is how they got the answer $k = n+2$. The way I would go about it is by completing squares: $n^2+6n = n^2 + 6n + 9 - 9 = (n+3)^2$, so $k$ is close to, but cannot exceed $n+3$. – player3236 Oct 22 '20 at 13:13
  • Ok,but then I am left with k and n ,do I replace k with [n^2+6n]? Because then I would get the integer part of x squared for which I cannot do nothing to.... – ROBOTICS Oct 22 '20 at 13:19
  • From $k^2 \le n^2 + 6n < (k+1)^2$ we can take square roots to obtain $k \le \sqrt {n^2+6n} < k+1$. This satisfies the definition of $[n^2+6n]$. – player3236 Oct 22 '20 at 13:22
  • Ok I now understand the question ,thank you very much! – ROBOTICS Oct 22 '20 at 13:23

1 Answers1

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The integer part of the square root is what you would get if you performed the root operation normally and then chopped off the decimal digits, leaving behind an integer.

For $n\ge2$, $$(n+2)^2=n^2+4n+4\le n^2+6n<n^2+6n+9=(n+3)^2$$ So the integer part is $n+2$ if $n\ge2$. (For $n=1$ the integer part is $2$.)

Parcly Taxel
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