Which of the following equations has exactly the same solutions as (x-7)(x+9)=0 ? x^2+2x=-63 x^2-2x-63=0 x^2=-2x+63 x^2=2x-63

To find which of the given equations has exactly the same solutions as \((x - 7)(x + 9) = 0\), we first need to identify the solutions of the original equation.

The equation \((x - 7)(x + 9) = 0\) has solutions when either factor is equal to zero:

1. \(x - 7 = 0 \implies x = 7\)
2. \(x + 9 = 0 \implies x = -9\)

So the solutions are \(x = 7\) and \(x = -9\).

Now let's analyze each of the provided options one by one to see which one produces the same solutions.

1. Option: \(x^2 + 2x = -63\)

   Rearranging gives: \[ x^2 + 2x + 63 = 0 \]

   Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

   Here, \(a = 1\), \(b = 2\), \(c = 63\): \[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 63}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 - 252}}{2} \] The discriminant \(4 - 252 = -248\) is negative, which means this equation has no real solutions.

2. Option: \(x^2 - 2x - 63 = 0\)

   Using the quadratic formula: Here, \(a = 1\), \(b = -2\), \(c = -63\): \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-63)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 252}}{2} = \frac{2 \pm \sqrt{256}}{2} = \frac{2 \pm 16}{2} \] This gives us: \[ x = \frac{18}{2} = 9 \quad \text{and} \quad x = \frac{-14}{2} = -7 \] The solutions are \(x = 9\) and \(x = -7\), which are not the same as the original solutions.

3. Option: \(x^2 = -2x + 63\)

   Rearranging gives: \[ x^2 + 2x - 63 = 0 \]

   Using the quadratic formula again: Here, \(a = 1\), \(b = 2\), \(c = -63\): \[ x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-63)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 252}}{2} = \frac{-2 \pm \sqrt{256}}{2} = \frac{-2 \pm 16}{2} \] This gives us: \[ x = \frac{14}{2} = 7 \quad \text{and} \quad x = \frac{-18}{2} = -9 \] The solutions are \(x = 7\) and \(x = -9\), which are the same as the original solutions.

4. Option: \(x^2 = 2x - 63\)

   Rearranging gives: \[ x^2 - 2x + 63 = 0 \]

   Using the quadratic formula: Here, \(a = 1\), \(b = -2\), \(c = 63\): \[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 63}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 252}}{2} \] The discriminant is again negative (\(4 - 252 = -248\)), meaning no real solutions.

After analyzing all options, the equation that has exactly the same solutions as \((x - 7)(x + 9) = 0\) is:

\(x^2 = -2x + 63\).