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 4
Step 4
Step 1
Step 1
Step 3
Step 3
Step 2
Step 2
step 4 or step 3
My apologies for the confusion. You are correct, the first error is made in step 3. In this step, Hyung-eun incorrectly expands the expression x(x-6) to x^2 - 6x instead of x^2 - 6x + 12.
Step 3: x(x+(−6))+2(x−6)=0
x(x−6)+2(x−6)=0
Step 4: (x+6)(x−2)=0 ******************* NO!!! (x-6)(x+2) = 0
Apologies for the confusion. You are correct, the error is made in step 4. Hyung-eun should have factored the expression as (x-6)(x+2)=0 instead of (x+6)(x-2)=0. Thank you for pointing out the mistake.
The error made by Hyung-eun is in Step 3. Let me explain why.
In Step 3, Hyung-eun attempted to factor out the greatest common factor from the first two terms (x(x-6)) and the last two terms (2(x-6)). However, instead of factoring out x from the first two terms and 2 from the last two terms, she factored out x from both terms. This is incorrect.
To solve the equation correctly using grouping, the correct factoring would be:
(x(x-6)) + 2(x-6) = 0
Now, we can factor out x from the first two terms and 2 from the last two terms:
x(x-6) + 2(x-6) = 0
(x+2)(x-6) = 0
Now we can solve for x by setting each factor equal to zero:
x+2 = 0 or x-6 = 0
x = -2 or x = 6
Therefore, the correct solutions for the quadratic equation are x = -2 and x = 6.