To solve the equation \( x + 4x - 5 = 7 \) using the zero product property, we first need to simplify the equation and bring all terms to one side.
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Simplify the equation: \[ x + 4x - 5 = 7 \] Combine like terms: \[ 5x - 5 = 7 \]
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Move all terms to one side: \[ 5x - 5 - 7 = 0 \] Simplifying this gives: \[ 5x - 12 = 0 \]
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Rearrange: \[ 5x = 12 \]
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Divide by 5: \[ x = \frac{12}{5} \]
The above steps lead to a linear equation rather than a quadratic one, and thus, we do not apply the zero product property here since it is specifically used for equations of the form \( ax^2 + bx + c = 0 \).
If we were to mistakenly treat it as a quadratic, we might consider the product of factors, but since the resulting equation is linear, we cannot factor \( 5x - 12 \) into the form of two products that equal zero.
Evaluating Student Solutions:
Let's review the students' work practices:
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Correct Student Solution: If Jamal correctly simplified and found \( x = \frac{12}{5} \), he has accurately solved a linear equation. He knows the steps of simplifying the equation correctly and has worked through the math.
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Incorrect Student Solution: If George wrote the equation as \( x + 4x - 5 = 0 \) or tried to factor it for a zero product property without ensuring it was a quadratic equation set to 0, then his solution is inaccurate. The equation he ended with was not a quadratic form that could be factored as a product of two factors.
Conclusion:
In this case, we conclude that Jamal's method was correct because he followed through the proper steps to solve a linear equation, while George's misunderstanding of what qualifies as a quadratic and incorrect attempt to use the zero product property resulted in an inaccurate solution. To ensure correctness, students should understand the types of equations they are dealing with before applying specific methods like factorizations or the zero product property.
If either student had the quadratic form \( (x - p)(x - q) = 0 \), they could have applied the zero product property. However, this is not applicable here as we are handling a linear equation.
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