Let's carefully analyze the problem Cyrus worked through:
1. Starting Equation:
[tex]\[
\frac{1}{3}(x-1)^2 = 12
\][/tex]
To eliminate the fraction, multiply both sides of the equation by 3:
[tex]\[
(x-1)^2 = 36 \quad \text{(Step 1)}
\][/tex]
This step is correct.
2. Taking the Square Root:
The next step is where Cyrus encounters an error. When solving [tex]\((x-1)^2 = 36\)[/tex], we must consider both the positive and negative square roots:
[tex]\[
x - 1 = 6 \quad \text{or} \quad x - 1 = -6
\][/tex]
However, Cyrus wrote only:
[tex]\[
x - 1 = 6 \quad \text{(Step 2)}
\][/tex]
Missing the second potential solution [tex]\((x - 1 = -6)\)[/tex] is the mistake.
3. Solving for [tex]\( x \)[/tex]:
From [tex]\(x - 1 = 6\)[/tex]:
[tex]\[
x = 6 + 1 = 7 \quad \text{(Step 3)}
\][/tex]
This solution is correct but incomplete.
4. Finding the Second Solution:
We also need to consider:
[tex]\[
x - 1 = -6
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
x = -6 + 1 = -5
\][/tex]
Thus, the two solutions to the equation [tex]\(\frac{1}{3}(x-1)^2 = 12\)[/tex] are:
[tex]\[
x = 7 \quad \text{or} \quad x = -5
\][/tex]
Error Identification:
- Cyrus made an error in Step 2 by not considering both potential solutions [tex]\( x - 1 = 6 \)[/tex] and [tex]\( x - 1 = -6 \)[/tex].
To summarize, the error occurred in Step 2 because Cyrus did not account for both the positive and negative square roots when solving the equation. The correct solutions are [tex]\( x = 7 \)[/tex] and [tex]\( x = -5 \)[/tex].
