To determine which of the given equations has the same solution as \((x-7)(x+9)=0\), we first need to solve the original equation.
The solutions to \((x-7)(x+9)=0\) can be found by setting each factor equal to zero:
- \(x - 7 = 0 \Rightarrow x = 7\)
- \(x + 9 = 0 \Rightarrow x = -9\)
So, the solutions are \(x = 7\) and \(x = -9\).
Next, we will check each of the options to see if they also yield these same solutions.
Option A: \(x^2 = 2x - 63\)
Rearranging gives: \[ x^2 - 2x + 63 = 0 \] Using the quadratic formula to find the solutions: \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 63}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 252}}{2} = \frac{2 \pm \sqrt{-248}}{2} \] Since the discriminant is negative, this does not have real solutions.
Option B: \(x^2 = -2x + 63\)
Rearranging gives: \[ x^2 + 2x - 63 = 0 \] Using the quadratic formula: \[ x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-63)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 252}}{2} = \frac{-2 \pm \sqrt{256}}{2} = \frac{-2 \pm 16}{2} \] Calculating:
- \(x = \frac{14}{2} = 7\)
- \(x = \frac{-18}{2} = -9\)
This option has the same solutions \(x = 7\) and \(x = -9\).
Option C: \(x^2 + 2x = -63\)
Rearranging gives: \[ x^2 + 2x + 63 = 0 \] Using the quadratic formula: \[ x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot 63}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 - 252}}{2} = \frac{-2 \pm \sqrt{-248}}{2} \] This again yields no real solutions.
Option D: \(x^2 - 2x - 63 = 0\)
Using the quadratic formula: \[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-63)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 252}}{2} = \frac{2 \pm \sqrt{256}}{2} = \frac{2 \pm 16}{2} \] Calculating:
- \(x = \frac{18}{2} = 9\)
- \(x = \frac{-14}{2} = -7\)
These solutions are \(x = 9\) and \(x = -7\), which do not match.
In conclusion, the equation that has exactly the same solutions as \((x-7)(x+9)=0\) is option B: \(x^2 = -2x + 63\).