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

Question

x
2
−
4
x
−
12
=
0
by grouping. Her work is shown below. At which step does Hyung-eun first make an error?

x2−4x−12=0
x
2
−
4
x
−
12
=
0

Step1: x2+(−6x)+2x−12=0
x
2
+
(
−
6
x
)
+
2
x
−
12
=
0

Step 2: (x2+(−6x))+(2x−12)=0
(
x
2
+
(
−
6
x
)
)
+
(
2
x
−
12
)
=
0

Step 3: x(x+(−6))+2(x−6)=0
x
(
x
+
(
−
6
)
)
+
2
(
x
−
6
)
=
0

x(x−6)+2(x−6)=0
x
(
x
−
6
)
+
2
(
x
−
6
)
=
0

Step 4: (x+6)(x−2)=0
(
x
+
6
)
(
x
−
2
)
=
0

Step 5: x+6=0
x
+
6
=
0
or x−2=0
x
−
2
=
0

x=−6
x
=
−
6
or x=2
x
=
2

(1 point)
Responses

Step 4
Step 4

Step 3
Step 3

Step 1
Step 1

Step 2
Step 2
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Item
1
, Fully Attempted.
Item
2
, Fully Attempted.
Item
3
, Fully Attempted.
Item
4
, Unattempted.
Item
5
, Unattempted.

Answer 1

Let's analyze Hyung-eun's steps in solving the quadratic equation \( x^2 - 4x - 12 = 0 \) by grouping.

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

2. Step 1:
   She rewrites it as \( x^2 + (-6x) + 2x - 12 = 0 \).
   This step looks accurate; however, the grouping itself is incorrect. The correct grouping to factor would involve products that equal -12 and sum to -4, which would be -6 and +2. Hydung-eun's grouping appears to be aiming to break down -4x incorrectly.

3. Step 2:
   Grouping becomes: \( (x^2 + (-6x)) + (2x - 12) = 0 \).
   This step is also correct as per her grouping. But again, the phrasing for choosing -6x with 2x for grouping doesn't lead to the correct factorization.

4. Step 3:
   She factors out the common terms: \( x(x + (-6)) + 2(x - 6) = 0 \).
   This is where the error starts to manifest because of how she factored the previous terms. The correct factorization from the step before would lead to \( (x - 6) + 2(x + 6) \). Since she factored incorrectly, the next steps will be affected.

5. Step 4:
   She writes \( (x + 6)(x - 2) = 0 \).
   This is incorrect as a result of the incorrect factoring in the previous step. She should've had a different pairing leading to correct factors.

6. Step 5:
   Solving gives \( x + 6 = 0 \) or \( x - 2 = 0 \), leading to \( x = -6 \) or \( x = 2 \), which is not even the correct pairs and terms originating from the original equation.

Conclusion

The first significant error occurs in Step 1, where the quadratic is incorrectly grouped. The correct grouping of \( -4x \) should capitalize on the sum-product relationship leading to the right pairs for factorization.

So, the answer is Step 1 as the first place where an error is made.