Question

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

Answer 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.
\[
\begin{aligned}
&v+2=3 \\
&v+2=-3
\end{aligned}
\]
\(\mathbf{8}\) Solve the 1 st equation: \(v+2=3\).
\[
v=1
\]
9 Solve the 2 nd equation: \(v+2=-3\).
\[
v=-5
\]
10 Collect all solutions.
\[
v=1,-5
\]