Solving Quadratic Equations Using Common Factors Quick Check

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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

(1 point)
Responses

Step 3
Step 3

Step 2
Step 2

Step 1
Step 1

Step 4
Step 4
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Answer 1

Let's analyze Hyung-eun's steps:

1. The original equation is \( x^2 - 4x - 12 = 0 \).

2. Step 1: She rewrites the equation as \( x^2 + (-6x) + 2x - 12 = 0 \). The correct factoring should give two terms that combine to give \( -4x \) when summed. Here, she splits \( -4x \) into \( (-6x) + (2x) \), which is valid.

3. Step 2: She groups the terms as \( (x^2 + (-6x)) + (2x - 12) = 0 \). This step is still correct as she is grouping the terms.

4. Step 3: She factored it as \( x(x + (-6)) + 2(x - 6) = 0 \). However, the term \( 2(x - 6) \) is incorrect. The correct split from the previous grouping should yield \( 2(x + 6) \).

At this point, the error occurs in Step 3. The step should have involved factoring \( 2(x + 6) \) instead of \( 2(x - 6) \).

Thus, the first error occurs at Step 3.