Problem of the Week

Updated at Sep 6, 2021 5:25 PM

This week's problem comes from the equation category.

How would you solve the equation \(\frac{{(v+2)}^{2}-6}{3}=1\)?

Let's begin!



\[\frac{{(v+2)}^{2}-6}{3}=1\]

1
Multiply both sides by \(3\).
\[{(v+2)}^{2}-6=1\times 3\]

2
Simplify  \(1\times 3\)  to  \(3\).
\[{(v+2)}^{2}-6=3\]

3
Add \(6\) to both sides.
\[{(v+2)}^{2}=3+6\]

4
Simplify  \(3+6\)  to  \(9\).
\[{(v+2)}^{2}=9\]

5
Take the square root of both sides.
\[v+2=\pm \sqrt{9}\]

6
Since \(3\times 3=9\), the square root of \(9\) is \(3\).
\[v+2=\pm 3\]

7
Break down the problem into these 2 equations.
\[v+2=3\]
\[v+2=-3\]

8
Solve the 1st equation: \(v+2=3\).
\[v=1\]

9
Solve the 2nd equation: \(v+2=-3\).
\[v=-5\]

10
Collect all solutions.
\[v=1,-5\]

Done
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