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# HELLLLP! QUADRATIC INEQUALITIES! WORD PROBLEMS!!

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A band is marching in a rectangular formation with dimensions $n-2$ and $n + 8$. In the second stage of their performance, they re-arrange to form a different rectangle with dimensions $n$ and $2n - 3$, excluding all the drummers. If there are at least 4 drummers, then find the sum of all possible values of $n$.

Mar 14, 2019

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$$\text{There are }M=(n-2)(n+8) \text{ total marchers}\\ \text{let }D\text{ be the number of drummers},~D\geq 4\\ M-D = n(2n-3)$$

$$n^2+6n-16 = 2n^2 - 3n+D\\ n^2 -9n+(16+D)=0\\ n = \dfrac{9\pm \sqrt{81-4(16+D)}}{2} \in \mathbb{N}$$

$$81 - 4(16+D) = k^2,~k \in \mathbb{N}\\ \text{The only possible value for }D \text{ is 4}\\ n = \dfrac{9 \pm 1}{2} = 5,4\\ 4+5=9$$

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Mar 14, 2019

#1
+6196
+2

$$\text{There are }M=(n-2)(n+8) \text{ total marchers}\\ \text{let }D\text{ be the number of drummers},~D\geq 4\\ M-D = n(2n-3)$$

$$n^2+6n-16 = 2n^2 - 3n+D\\ n^2 -9n+(16+D)=0\\ n = \dfrac{9\pm \sqrt{81-4(16+D)}}{2} \in \mathbb{N}$$

$$81 - 4(16+D) = k^2,~k \in \mathbb{N}\\ \text{The only possible value for }D \text{ is 4}\\ n = \dfrac{9 \pm 1}{2} = 5,4\\ 4+5=9$$

Rom Mar 14, 2019