Question

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

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

(2 points)
0 / 10000 Word Limit
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Answer 1

To determine which student is incorrect, we need to analyze the steps taken by both Jamal and George in solving the quadratic equation \( x^2 + 4x - 5 = 7 \).

### Step 1: Set the Equation to Zero
Initially, we need to move all terms to one side of the equation to set it equal to zero.
Starting with \( x^2 + 4x - 5 = 7 \):

\[
x^2 + 4x - 5 - 7 = 0
\]

This simplifies to:

\[
x^2 + 4x - 12 = 0
\]

Both Jamal and George correctly set the equation to zero in Step 1.

### Step 2: Factor the Quadratic Equation
Now let's look at the factoring of the quadratic \( x^2 + 4x - 12 \).

**Jamal's factors:**
Jamal factored the equation as:

\[
(x + 6)(x - 2) = 0
\]

**George's factors:**
George factored the equation as:

\[
(x + 5)(x - 1) = 7
\]

Here, George's factorization is incorrect for the equation \( x^2 + 4x - 12 = 0 \) because \( (x + 5)(x - 1) \) does not equal \( x^2 + 4x - 12 \).

### Step 3: Setting Each Factor to Zero
Jamal set each factor correctly equal to 0:

\[
x + 6 = 0 \quad \Rightarrow \quad x = -6
\]
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]

George's equations are based on his incorrect factorization:

\[
x + 5 = 0 \quad \Rightarrow \quad x = -5
\]
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]

### Conclusion
Jamal is correct, and his solutions are \( x = -6 \) and \( x = 2\).

George is incorrect because his factorization \( (x + 5)(x - 1) \) does not yield \( x^2 + 4x - 12 = 0 \). The correct factors should lead to a factoring that satisfies the original quadratic equation correctly, and hence George's solutions of \( x = -5 \) and \( x = 1 \) are also incorrect.

Therefore, **George is the student who is incorrect.**