Table is provided for the quadratic equation. Oliver completed his work. Is he accurate?

Question

(x+3) (x+4) =0
x-3=0 x+4=0
x=3 x=4
1. Yes, his work is accurate.
2. No, he should have a third column where he should set x=0
3. No, x-3=0 in the second row should be x+3=0
4. No, x-4=0 in the second row should be x+4=0

Answer 1

Let's analyze Oliver's work step by step.

The original quadratic equation is:

\[ (x + 3)(x + 4) = 0 \]

To solve for \(x\), we set each factor equal to zero:

1. \(x + 3 = 0\) → \(x = -3\)
2. \(x + 4 = 0\) → \(x = -4\)

Now, let's check Oliver's work. Here is the breakdown of what he wrote:

• He wrote:
  \(x - 3 = 0\) (This is incorrect since the factor should be \(x + 3 = 0\)) \(x + 4 = 0\) (This is correct)
• He found:
  \(x = 3\) (This is incorrect since it should be \(x = -3\))
  \(x = 4\) (This is correct but corresponds to the wrong factor)

From his work, we can identify the following issues:

1. He incorrectly set \(x - 3 = 0\) instead of \(x + 3 = 0\).
2. The values he found, \(x = 3\) and \(x = 4\), do not match the correct solutions, which are \(x = -3\) and \(x = -4\).

Thus, the correct option among those provided would be:

3. No, x - 3 = 0 in the second row should be x + 3 = 0.