To find out which of the given equations has exactly the same solutions as \((x−7)(x+9)=0\), we first determine the solutions to the original equation.
Setting \((x−7)(x+9)=0\) gives us the solutions:
- \(x - 7 = 0 \Rightarrow x = 7\)
- \(x + 9 = 0 \Rightarrow x = -9\)
Thus, the solutions to the equation \((x−7)(x+9)=0\) are \(x = 7\) and \(x = -9\).
Now, we will examine each of the given equations to see if they yield the same solutions.
-
\(x^2 = 2x - 63\):
Rearrange to standard form:
\[ x^2 - 2x + 63 = 0 \]
The solutions can be found using the quadratic formula, and they will not yield 7 or -9. -
\(x^2 = -2x + 63\):
Rearrange to standard form:
\[ x^2 + 2x - 63 = 0 \]
This can be factored:
\((x - 7)(x + 9) = 0\)
This equation has solutions \(x = 7\) and \(x = -9\). -
\(x^2 + 2x = -63\):
Rearrange to standard form:
\[ x^2 + 2x + 63 = 0 \]
The solutions do not match 7 or -9. -
\(x^2 - 2x - 63 = 0\):
This factors into:
\((x - 9)(x + 7) = 0\)
The solutions will be \(x = 9\) and \(x = -7\), which do not match.
The only equation that has the exact same solutions \(x = 7\) and \(x = -9\) is:
\(x^2 + 2x - 63 = 0\).