Problem of the Week

Updated at Sep 17, 2018 5:18 PM

This week we have another equation problem:

How can we solve the equation \(\frac{{(2+n)}^{2}}{4(n-3)}=9\)?

Let's start!



\[\frac{{(2+n)}^{2}}{4(n-3)}=9\]

1
Multiply both sides by \(4(n-3)\).
\[{(2+n)}^{2}=9\times 4(n-3)\]

2
Simplify  \(9\times 4(n-3)\)  to  \(36(n-3)\).
\[{(2+n)}^{2}=36(n-3)\]

3
Expand.
\[4+4n+{n}^{2}=36n-108\]

4
Move all terms to one side.
\[4+4n+{n}^{2}-36n+108=0\]

5
Simplify  \(4+4n+{n}^{2}-36n+108\)  to  \(112-32n+{n}^{2}\).
\[112-32n+{n}^{2}=0\]

6
Factor \(112-32n+{n}^{2}\).
\[(n-28)(n-4)=0\]

7
Solve for \(n\).
\[n=28,4\]

Done
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