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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 4
Step 4

Step 3
Step 3

Step 2
Step 2
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1 answer

To determine where Hyung-eun first makes an error, let's examine each step of her work.

  1. The original equation is \(x^2 - 4x - 12 = 0\).
  2. In Step 1, she rewrites the equation as \(x^2 + (-6x) + 2x - 12 = 0\). This grouping is incorrect because the terms \( -4x \) cannot be split into \( -6x \) and \( 2x \) while maintaining equality. The correct way to group should have kept the coefficients adding up to \(-4\).

This shows that Step 1 contains the first error.

  1. Moving to Step 2, she groups terms as \( (x^2 + (-6x)) + (2x - 12) = 0 \), which is based on her incorrect grouping from the previous step.
  2. Step 3 factors out \( x \) and \( 2 \), leading to \( x(x - 6) + 2(x - 6) = 0\). This could actually lead to the correct product form if Step 1 were correctly grouped, but it is still based on Step 1's error.
  3. Step 4 factors as \( (x + 6)(x - 2) = 0\). Here, she made a mistake because \( x(x - 6) + 2(x - 6) \) should indeed factor to \( (x - 6)(x + 2) \).
  4. Step 5 derives \( x + 6 = 0 \) or \( x - 2 = 0 \), leading to incorrect solutions due to earlier missteps.

Thus, the first error occurs in Step 1.

The answer is: Step 1.