Question

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Use the tables to answer the question.

When trying to solve the quadratic equation x2+4x−5=7
, Jamal and George each provided their work in the tables.

Jamal

Step 1: Set equation = 0 x2+4x−12=0

Step 2: Write quadratic equation as the product of two factors (x+6)(x−2)=0

Step 3: Set each factor equal to 0 x+6=0 and x−2=0

Step 4: Solve each equation x=−6 and x=2



George

Step 1: Set equation = 0 x2+4x−5=7

Step 2: Write quadratic equation as the product of two factors (x+5)(x−1)=7

Step 3: Set each factor equal to 0 x+5=0 and x−1=0

Step 4: Solve each equation x=−5 and x=1

Question: Which student is incorrect? Explain how you know that his solution is inaccurate.

(2 points)

Answer 1

To determine which student is incorrect, let's first verify the work of both Jamal and George.

### Jamal's Work:
1. **Set equation = 0**:
- He starts with the equation \( x^2 + 4x - 5 = 7 \).
- Subtracting 7 from both sides gives:
\[
x^2 + 4x - 12 = 0.
\]

2. **Write quadratic equation as the product of two factors**:
- Jamal factored it as \( (x + 6)(x - 2) = 0 \).

3. **Set each factor equal to 0**:
- He correctly sets \( x + 6 = 0 \) and \( x - 2 = 0 \).

4. **Solve each equation**:
- Solving gives \( x = -6 \) and \( x = 2 \).

### George's Work:
1. **Set equation = 0**:
- He also starts with \( x^2 + 4x - 5 = 7 \) and similarly moves to:
\[
x^2 + 4x - 12 = 0.
\]

2. **Write quadratic equation as the product of two factors**:
- George factors it as \( (x + 5)(x - 1) = 7 \), which is incorrect since he has not correctly factored the left-hand side.

3. **Set each factor equal to 0**:
- He sets \( x + 5 = 0 \) and \( x - 1 = 0 \).

4. **Solve each equation**:
- Solving gives \( x = -5 \) and \( x = 1 \).

### Conclusion:
George is incorrect. The crucial error lies in his Step 2 where he factored the quadratic equation incorrectly. The correct factoring from \( x^2 + 4x - 12 = 0 \) should yield \( (x + 6)(x - 2) \), not \( (x + 5)(x - 1) \). Therefore, his solutions \( x = -5 \) and \( x = 1 \) are not valid for the given equation. Jamal's solutions, \( x = -6 \) and \( x = 2 \), are correct.