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)

1 answer

To determine which student is incorrect, we need to analyze their steps in solving the equation \( x^2 + 4x - 5 = 7 \).

  1. First, we need to set the given equation to zero: \[ x^2 + 4x - 5 - 7 = 0 \implies x^2 + 4x - 12 = 0 \] Both students correctly set the equation to zero.

  2. Next, we should factor the quadratic equation \( x^2 + 4x - 12 = 0 \). To find the correct factors, we need two numbers that multiply to \(-12\) (the constant term) and add up to \(4\) (the coefficient of \(x\)). The correct factors are \(6\) and \(-2\): \[ (x + 6)(x - 2) = 0 \] This matches Jamal's factorization.

    On the other hand, George incorrectly factored the equation as: \[ (x + 5)(x - 1) = 0 \] The product \( (x + 5)(x - 1) \) expands to \( x^2 + 4x - 5 \), which does not equal \( x^2 + 4x - 12 \).

  3. In Step 3, both students set their factors equal to zero. Jamal's factors \( x + 6 = 0 \) and \( x - 2 = 0 \) yield: \[ x = -6 \quad \text{and} \quad x = 2 \] George's incorrect factors \( x + 5 = 0 \) and \( x - 1 = 0 \) yield: \[ x = -5 \quad \text{and} \quad x = 1 \]

Since George made an error in his factorization and derived incorrect roots, he is the student who is incorrect.

In summary, George's solution is inaccurate because he improperly factored the equation, leading to incorrect roots. The correct roots from Jamal's work are \( x = -6 \) and \( x = 2 \).

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