Question

Hyung-eun is solving the quadratic equation x2−4x−12=0 by grouping. Her work is shown below. At which step does Hyung-eun first make an error? x2−4x−12=0 Step1: x2+(−6x)+2x−12=0 Step 2: (x2+(−6x))+(2x−12)=0 Step 3: x(x+(−6))+2(x−6)=0 x(x−6)+2(x−6)=0 Step 4: (x+6)(x−2)=0 Step 5: x+6=0 or x−2=0 x=−6 or x=2

Answer 1

To determine where Hyung-eun first makes an error in solving the quadratic equation \( x^2 - 4x - 12 = 0 \) using grouping, let's analyze each step provided:

1. **Step 1**: \( x^2 + (-6x) + 2x - 12 = 0 \)
- This step involves rewriting \(-4x\) as \((-6x + 2x)\). This is correct since \(-6 + 2 = -4\).

2. **Step 2**: \( (x^2 + (-6x)) + (2x - 12) = 0 \)
- This groups the first two terms and the last two correctly. So far, this step is also correct.

3. **Step 3**: \( x(x + (-6)) + 2(x - 6) = 0 \)
- In this grouping, \( x^2 + (-6x) \) is factored correctly as \( x(x - 6) \), but \( 2x - 12 \) should be factored as \( 2(x - 6) \). Hyung-eun has still followed the steps correctly to this point.

4. **Step 4**: \( (x + 6)(x - 2) = 0 \)
- This step is where the first error occurs. The correct factorization from Step 3 should yield:
\[
x(x - 6) + 2(x - 6) = (x + 2)(x - 6) = 0
\]
Instead, Hyung-eun wrote \((x + 6)(x - 2)\), which is incorrect.

5. **Step 5**: \( x + 6 = 0 \text{ or } x - 2 = 0 \)
- This step follows incorrectly from Step 4's factorization error.

In conclusion, the first error occurs in **Step 4**, where the factors were misidentified as \((x + 6)(x - 2)\) instead of the correct factors \( (x + 2)(x - 6) \).