Problem of the Week

Updated at Nov 11, 2019 2:40 PM

This week's problem comes from the equation category.

How would you solve the equation \(6+{(3(3-z))}^{2}=42\)?

Let's begin!



\[6+{(3(3-z))}^{2}=42\]

1
Use Multiplication Distributive Property: \({(xy)}^{a}={x}^{a}{y}^{a}\).
\[6+{3}^{2}{(3-z)}^{2}=42\]

2
Simplify  \({3}^{2}\)  to  \(9\).
\[6+9{(3-z)}^{2}=42\]

3
Subtract \(6\) from both sides.
\[9{(3-z)}^{2}=42-6\]

4
Simplify  \(42-6\)  to  \(36\).
\[9{(3-z)}^{2}=36\]

5
Divide both sides by \(9\).
\[{(3-z)}^{2}=\frac{36}{9}\]

6
Simplify  \(\frac{36}{9}\)  to  \(4\).
\[{(3-z)}^{2}=4\]

7
Take the square root of both sides.
\[3-z=\pm \sqrt{4}\]

8
Since \(2\times 2=4\), the square root of \(4\) is \(2\).
\[3-z=\pm 2\]

9
Break down the problem into these 2 equations.
\[3-z=2\]
\[3-z=-2\]

10
Solve the 1st equation: \(3-z=2\).
\[z=1\]

11
Solve the 2nd equation: \(3-z=-2\).
\[z=5\]

12
Collect all solutions.
\[z=1,5\]

Done
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