Hyung-eun is solving the quadratic equation x2−4x−12=0

Question

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 4
Step 4

Step 2

Answer 1

Let's go through Hyung-eun's work step by step to identify the first error:

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

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

   • This is where the error begins. The expression \(-4x\) should be split into two terms that add up to \(-4x\) and multiply to \(-12\). The correct grouping should be \( -6x + 2x \) which is okay, but since we are supposed to factor \( x^2 - 4x - 12 \) directly, it's generally not split in this manner when grouping.

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

   • This is just a rearrangement of Step 1, and the grouping is still incorrect; thus, it's already problematic.

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

   • Here, she factored \( x^2 - 6x + 2x - 12 \), but the factorization of \( x^2 - 4x - 12 \) should be recognized from the first instance. The grouping is off-balance.

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

   • This factorization is not correct based on the original expression as it should have been \( (x-6)(x+2) = 0 \).

Conclusion: The first clear error happens in Step 1 when she incorrectly splits \(-4x\) into \(-6x + 2x\). Therefore, the answer is:

Step 1