Question

Hyung is solving the quadratic equation x2−4x−12=0x2−4x−12=0 by grouping. Her work is shown below. At which step does Hyung-eun first make an error? x2−4x−12=0x2−4x−12=0 Step 1: x2+(−6x)+2x−12=0x2+(−6x)+2x−12=0 Step 2: (x2+(−6x))+(2x−12)=0(x2+(−6x))+(2x−12)=0 Step: 3: x(x−6)+2(x−6)=0x(x−6)+2(x−6)=0 Step 4: (x+6)(x−2)=0(x+6)(x−2)=0 Step 5: x+6=0 or x−2=0x+6=0 or x−2=0 x=-6 or x=2 (1 point) Responses Step 3 Step 3 Step 4 Step 4 Step 1 Step 1 Step 2

Answer 1

To identify the first error in Hyung's work while solving the quadratic equation \( x^2 - 4x - 12 = 0 \) by grouping, let's review each step:

1. Step 1: \( x^2 + (-6x) + 2x - 12 = 0 \)

   • Here, Hyung is attempting to split \( -4x \) into \( -6x + 2x \), which is correct since \(-6 + 2 = -4\).

2. Step 2: \( (x^2 + (-6x)) + (2x - 12) = 0 \)

   • This step correctly groups the terms as intended.

3. Step 3: \( x(x - 6) + 2(x - 6) = 0 \)

   • In this step, she factored out \( x \) from \( x^2 - 6x \) and \( 2 \) from \( 2x - 12 \), which is also correct. Both terms indeed have \( (x - 6) \), so this part is accurate.

4. Step 4: \( (x + 6)(x - 2) = 0 \)

   • Here comes the error. From the previous step, the correct expression is \( (x - 6)(x + 2) = 0 \). This was incorrectly factored as \( (x + 6)(x - 2) \).

5. Step 5: \( x + 6 = 0 \) or \( x - 2 = 0 \)

   • This follows from Step 4, but since Step 4 itself is wrong, the solutions derived ( \( x = -6 \) or \( x = 2 \)) are also incorrect.

Thus, the first error occurs in Step 4.