Question

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 1
Step 1

Step 3
Step 3

Step 2
Step 2

Step 4
Step 4
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Answer 1

To determine where Hyung-eun made an error in her work, let's analyze each step critically.

Original equation: \( x^2 - 4x - 12 = 0 \)

Step 1: Hyung-eun rewrote the middle term, \(-4x\), as \((-6x) + (2x)\): \[ x^2 + (-6x) + 2x - 12 = 0 \] This step is correct because \(-6 + 2 = -4\).

Step 2: She groups the first two terms and the last two terms: \[ (x^2 + (-6x)) + (2x - 12) = 0 \] This step is also correct.

Step 3: Here, she factors out the common terms from each group: \[ x(x + (-6)) + 2(x - 6) = 0 \] This is where the first error occurs. The correct factorization should be: \[ (x(x - 6) + 2(x - 6) = 0 \quad \text{(not } 2(x - 6)\text{)} \]

So, the expression she should have had after factoring would be: \[ (x - 6)(x + 2) = 0 \]

Step 4: Hyung-eun writes: \((x + 6)(x - 2) = 0\) This is incorrect based on her earlier work because the factorization was not done correctly in Step 3.

Therefore, Hyung-eun first makes an error in Step 3.

Correct answer: Step 3.