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.
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.
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To determine which student's solution is incorrect, we need to evaluate their work against the original equation step by step.
The original equation is:
\[ x^2 + 4x - 5 = 7 \]
**Step 1: Set the equation to 0.**
To do this, we first subtract 7 from both sides:
\[ x^2 + 4x - 5 - 7 = 0 \]
This simplifies to:
\[ x^2 + 4x - 12 = 0 \]
**Step 2: Factor the quadratic equation.**
Now we need to factor \( x^2 + 4x - 12 \). We are looking for two numbers that multiply to -12 and add to 4. These numbers are 6 and -2, so the factored form is:
\[ (x + 6)(x - 2) = 0 \]
**Step 3: Set each factor equal to 0.**
We set each factor equal to zero:
1. \( x + 6 = 0 \) leads to \( x = -6 \)
2. \( x - 2 = 0 \) leads to \( x = 2 \)
**Step 4: Solve each equation.**
From the above step, we find the solutions:
\[ x = -6 \quad \text{and} \quad x = 2 \]
Now let's look at George's solution:
1. George incorrectly left the equation in the form \( x^2 + 4x - 5 = 7 \) without setting it to zero.
2. His factors \( (x + 5)(x - 1) = 7 \) are not correct for the equation \( x^2 + 4x - 12 = 0 \). The correct factors should lead to the left-hand side being equal to zero, not seven.
3. When he sets each factor equal to zero, he solves \( (x + 5) = 0 \) which yields \( x = -5 \) and \( (x - 1) = 0 \) which yields \( x = 1 \), neither of which are correct.
Based on this evaluation, **George is the incorrect student** because he did not properly set the equation to zero before factoring it, resulting in incorrect factors and solutions. Jamal's method and final answers of \( x = -6 \) and \( x = 2 \) are accurate.
The original equation is:
\[ x^2 + 4x - 5 = 7 \]
**Step 1: Set the equation to 0.**
To do this, we first subtract 7 from both sides:
\[ x^2 + 4x - 5 - 7 = 0 \]
This simplifies to:
\[ x^2 + 4x - 12 = 0 \]
**Step 2: Factor the quadratic equation.**
Now we need to factor \( x^2 + 4x - 12 \). We are looking for two numbers that multiply to -12 and add to 4. These numbers are 6 and -2, so the factored form is:
\[ (x + 6)(x - 2) = 0 \]
**Step 3: Set each factor equal to 0.**
We set each factor equal to zero:
1. \( x + 6 = 0 \) leads to \( x = -6 \)
2. \( x - 2 = 0 \) leads to \( x = 2 \)
**Step 4: Solve each equation.**
From the above step, we find the solutions:
\[ x = -6 \quad \text{and} \quad x = 2 \]
Now let's look at George's solution:
1. George incorrectly left the equation in the form \( x^2 + 4x - 5 = 7 \) without setting it to zero.
2. His factors \( (x + 5)(x - 1) = 7 \) are not correct for the equation \( x^2 + 4x - 12 = 0 \). The correct factors should lead to the left-hand side being equal to zero, not seven.
3. When he sets each factor equal to zero, he solves \( (x + 5) = 0 \) which yields \( x = -5 \) and \( (x - 1) = 0 \) which yields \( x = 1 \), neither of which are correct.
Based on this evaluation, **George is the incorrect student** because he did not properly set the equation to zero before factoring it, resulting in incorrect factors and solutions. Jamal's method and final answers of \( x = -6 \) and \( x = 2 \) are accurate.
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