Carla and Jordan were trying to solve the equation:

Question

\[(x+1)(x-1)=8\]
Carla said, “I'll multiply
\[(x+1)(x-1)\] and rewrite the equation as
\[x^2-1=8\]. Then I'll add
\[1\] to both sides and take the square root."
Jordan said, “The left-hand side is factored, so I'll use the zero product property.”
Whose solution strategy would work?
Choose 1 answer:
Choose 1 answer:
(Choice A) Only Carla's
A
Only Carla's
(Choice B) Only Jordan's
B
Only Jordan's
(Choice C) Both
C
Both
(Choice D) Neither
D
Neither

Answer 1

Let's analyze both Carla's and Jordan's approaches to solving the equation \((x+1)(x-1) = 8\).

Carla's Approach:

1. She rewrites the equation by multiplying out the left side: \[ (x+1)(x-1) = x^2 - 1. \] This gives her \(x^2 - 1 = 8\).
2. She then adds 1 to both sides: \[ x^2 = 9. \]
3. Finally, she takes the square root of both sides: \[ x = 3 \quad \text{or} \quad x = -3. \]

Carla's solution is valid.

Jordan's Approach: Jordan states that he will use the zero-product property. However, in order to use the zero-product property, he would need to set the left-hand side equal to zero: \[ (x+1)(x-1) - 8 = 0 \Rightarrow (x+1)(x-1) - 8 = 0. \] This would require him to manipulate the equation a little further.

To correctly use the zero-product property, he would first need to move the 8 to the opposite side: \[ (x+1)(x-1) - 8 = 0. \]

Then, he would need to factor or set up the equation: \[ (x^2 - 1 - 8) = 0 \Rightarrow x^2 - 9 = 0 \Rightarrow (x-3)(x+3) = 0. \] This leads to the same solutions: \[ x = 3 \text{ or } x = -3. \]

So, Jordan’s approach also works but requires a little rearranging first.

Since both Carla's and Jordan's methods lead to the correct solutions, the answer is:

(Choice C) Both.