Problem of the Week

Updated at Nov 28, 2022 11:55 AM

This week's problem comes from the equation category.

How would you solve the equation \({(2+4(3-u))}^{2}=36\)?

Let's begin!



\[{(2+4(3-u))}^{2}=36\]

1
Factor out the common term \(2\).
\[{(2(1+2(3-u)))}^{2}=36\]

2
Use Multiplication Distributive Property: \({(xy)}^{a}={x}^{a}{y}^{a}\).
\[{2}^{2}{(1+2(3-u))}^{2}=36\]

3
Simplify  \({2}^{2}\)  to  \(4\).
\[4{(1+2(3-u))}^{2}=36\]

4
Divide both sides by \(4\).
\[{(1+2(3-u))}^{2}=\frac{36}{4}\]

5
Simplify  \(\frac{36}{4}\)  to  \(9\).
\[{(1+2(3-u))}^{2}=9\]

6
Take the square root of both sides.
\[1+2(3-u)=\pm \sqrt{9}\]

7
Since \(3\times 3=9\), the square root of \(9\) is \(3\).
\[1+2(3-u)=\pm 3\]

8
Break down the problem into these 2 equations.
\[1+2(3-u)=3\]
\[1+2(3-u)=-3\]

9
Solve the 1st equation: \(1+2(3-u)=3\).
\[u=2\]

10
Solve the 2nd equation: \(1+2(3-u)=-3\).
\[u=5\]

11
Collect all solutions.
\[u=2,5\]

Done
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