Question

Tyrianne solved a quadratic equation. Her work is shown below.
In which step did Tyrianne make an error?



\[\begin{aligned}
\dfrac{1}{2}(x+4)^2-3&=29
\\\\
\dfrac{1}{2}(x+4)^2&=32&\text{Step }1
\\\\
(x+4)^2&=16&\text{Step }2
\\\\
x+4&=\pm4&\text{Step }3
\\\\
x=0&\text{ or }x=-8&\text{Step }4
\end{aligned}\]
Choose 1 answer:
Choose 1 answer:
(Choice A)
\[\text{Step } 1\]
A
\[\text{Step } 1\]
(Choice B)
\[\text{Step } 2\]
B
\[\text{Step } 2\]
(Choice C)
\[\text{Step } 3\]
C
\[\text{Step } 3\]
(Choice D)
\[\text{Step } 4\]
D
\[\text{Step } 4\]

Answer 1

Let's analyze Tyrianne's solution step by step to identify any errors.

1. **Step 1:**
\[
\frac{1}{2}(x+4)^2 - 3 = 29
\]
Tyrianne correctly adds 3 to both sides:
\[
\frac{1}{2}(x+4)^2 = 32
\]

2. **Step 2:**
She then multiplies both sides by 2 to eliminate the fraction, leading to:
\[
(x+4)^2 = 64
\]
(This is the error—not \(16\), as she incorrectly wrote.)

3. **Step 3:**
Tyrianne takes the square root:
\[
x+4 = \pm 8
\]

4. **Step 4:**
Solving for \(x\), she finds:
\[
x = 4 \quad \text{or} \quad x = -12
\]

Based on our analysis, the error occurred in **Step 2** where she incorrectly stated \((x+4)^2 = 16\) instead of \((x+4)^2 = 64\).

Therefore, the correct choice is:
**(Choice B) Step 2**.