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?

Question

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 4
Step 4

Step 2
Step 2

Step 3
Step 3

Step 1
Step 1
Skip to navigation
page 16 of 16

Answer 1

To determine where Hyung-eun first makes an error in her solution to the quadratic equation \( x^2 - 4x - 12 = 0 \), let’s carefully analyze each step.

Starting with the original equation:

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

   Here, the rewriting of \( -4x \) as \( -6x + 2x \) is correct.

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

   This grouping is also correct.

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

   At this step, Hyung-eun properly factored the first group \( x^2 - 6x \) to \( x(x - 6) \) and the second group \( 2x - 12 \) to \( 2(x - 6) \), which is correct.

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

   In this step, she combines the factored forms correctly.

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

   Here is where the first error occurs. The factors should not have been \( (x + 6)(x - 2) \); they should be \( (x - 6)(x + 2) \). This mistake impacts the solutions found.

So, the first error occurs in:

Response: Step 4.