Problem of the Week

Updated at Sep 23, 2024 9:22 AM

How would you solve the equation \((\frac{{5}^{2}}{{(4(2+n))}^{2}})=\frac{1}{16}\)?

Below is the solution.



\[(\frac{{5}^{2}}{{(4(2+n))}^{2}})=\frac{1}{16}\]

1
Remove parentheses.
\[\frac{{5}^{2}}{{(4(2+n))}^{2}}=\frac{1}{16}\]

2
Simplify  \({5}^{2}\)  to  \(25\).
\[\frac{25}{{(4(2+n))}^{2}}=\frac{1}{16}\]

3
Use Multiplication Distributive Property: \({(xy)}^{a}={x}^{a}{y}^{a}\).
\[\frac{25}{{4}^{2}{(2+n)}^{2}}=\frac{1}{16}\]

4
Simplify  \({4}^{2}\)  to  \(16\).
\[\frac{25}{16{(2+n)}^{2}}=\frac{1}{16}\]

5
Multiply both sides by \(16{(2+n)}^{2}\).
\[25=\frac{1}{16}\times 16{(2+n)}^{2}\]

6
Cancel \(16\).
\[25={(2+n)}^{2}\]

7
Take the square root of both sides.
\[\pm \sqrt{25}=2+n\]

8
Since \(5\times 5=25\), the square root of \(25\) is \(5\).
\[\pm 5=2+n\]

9
Switch sides.
\[2+n=\pm 5\]

10
Break down the problem into these 2 equations.
\[2+n=5\]
\[2+n=-5\]

11
Solve the 1st equation: \(2+n=5\).
\[n=3\]

12
Solve the 2nd equation: \(2+n=-5\).
\[n=-7\]

13
Collect all solutions.
\[n=3,-7\]

Done
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