To find which equation has exactly the same solutions as \((x - 7)(x + 9) = 0\), we first determine the solutions of the given equation.
The equation \((x - 7)(x + 9) = 0\) has solutions when:
- \(x - 7 = 0 \implies x = 7\)
- \(x + 9 = 0 \implies x = -9\)
Thus, the solutions are \(x = 7\) and \(x = -9\).
Now, let's check each option to see if any produces the same solutions.
Option A: \(x^2 - 2x - 63 = 0\)
To find the roots, we can use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -2\), and \(c = -63\):
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-63)}}{2 \cdot 1} \] \[ = \frac{2 \pm \sqrt{4 + 252}}{2} \] \[ = \frac{2 \pm \sqrt{256}}{2} \] \[ = \frac{2 \pm 16}{2} \]
This gives:
\[ x = \frac{18}{2} = 9 \quad \text{and} \quad x = \frac{-14}{2} = -7 \]
The solutions are \(x = 9\) and \(x = -7\), which do not match.
Option B: \(x^2 + 2x = -63\)
Rearranging gives us:
\[ x^2 + 2x + 63 = 0 \]
Using the quadratic formula with \(a = 1\), \(b = 2\), \(c = 63\):
\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 63}}{2 \cdot 1} \] \[ = \frac{-2 \pm \sqrt{4 - 252}}{2} \] \[ = \frac{-2 \pm \sqrt{-248}}{2} \]
This indicates there are complex solutions, which do not match.
Option C: \(x^2 = 2x - 63\)
Rearranging gives us:
\[ x^2 - 2x + 63 = 0 \]
Finding roots:
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 63}}{2 \cdot 1} \] \[ = \frac{2 \pm \sqrt{4 - 252}}{2} \] \[ = \frac{2 \pm \sqrt{-248}}{2} \]
Again, this means complex solutions, not matching.
Option D: \(x^2 = -2x + 63\)
Rearranging gives us:
\[ x^2 + 2x - 63 = 0 \]
Using the quadratic formula:
\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-63)}}{2 \cdot 1} \] \[ = \frac{-2 \pm \sqrt{4 + 252}}{2} \] \[ = \frac{-2 \pm \sqrt{256}}{2} \] \[ = \frac{-2 \pm 16}{2} \]
This yields:
\[ x = \frac{14}{2} = 7 \quad \text{and} \quad x = \frac{-18}{2} = -9 \]
Thus, the solutions are \(x = 7\) and \(x = -9\), matching the original equation.
Therefore, the correct answer is:
D: \(x^2 = -2x + 63\).