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 4
Step 4

Step 1
Step 1

Step 3
Step 3

Step 2
Step 2
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Answer 1

Let's analyze Hyung-eun's steps to identify where the error occurs.

Starting with the original equation:

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

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

In Step 1, Hyung-eun attempts to break down \( -4x \) into \( -6x + 2x \). This step is actually correct because \( -6x + 2x = -4x \).

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

Grouping the terms as shown is also correct.

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

Here, it appears she factored \( x^2 - 6x \) correctly as \( x(x - 6) \) but made a mistake in her second grouping. The grouping should maintain consistency with the factored terms.

Following from \( 2(x - 6) \), it should actually be \( 2(x - 6) \). But when rewritten, we need to maintain that both groups provide the same constant term \( -12 \). There is confusion because \( -6 \) is not part of the factoring correctly applied when expressed with respect to the original quadratic form.

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

In this step, Hyung-eun has incorrectly factored the obtained expression \( x(x - 6) + 2(x - 6) \) as \( (x + 6)(x - 2) \) instead of \( (x + 6)(x - 2) \).

The error actually occurs in Step 4. The correct factoring from the previous expression \( x(x - 6) + 2(x - 6) \) should yield:

\[ (x + 2)(x - 6) = 0 \]

Thus the correct answers should be:

• The correct factors would yield solutions \( x = -2 \) and \( x = 6 \).

It follows that Hyung-eun makes an error in Step 4.