To find which of the given equations has the same solutions as \((x−7)(x+9)=0\), we first find the solutions to that equation.
\((x−7)(x+9)=0\) has solutions when:
- \(x - 7 = 0 \Rightarrow x = 7\)
- \(x + 9 = 0 \Rightarrow x = -9\)
So, the solutions to \((x−7)(x+9)=0\) are \(x = 7\) and \(x = -9\).
Now we will check each of the given equations to see if they yield these solutions.
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Equation: \(x^2 - 2x - 63 = 0\)
To factor this equation, we look for two numbers that multiply to \(-63\) and add to \(-2\). The numbers are \(7\) and \(-9\). Thus: \[ (x - 7)(x + 9) = 0 \] So the solutions are \(x = 7\) and \(x = -9\), which matches the solutions we found. -
Equation: \(x^2 = 2x - 63\)
Rearranging gives: \[ x^2 - 2x + 63 = 0 \] This quadratic does not factor in a way that gives the same \(x\) values as \(7\) and \(-9\). -
Equation: \(x^2 + 2x = -63\)
Rearranging gives: \[ x^2 + 2x + 63 = 0 \] The solutions to this quadratic do not match what we found. -
Equation: \(x^2 = -2x + 63\)
Rearranging gives: \[ x^2 + 2x - 63 = 0 \] This can be factored as: \[ (x - 7)(x + 9) = 0 \] Thus, the solutions are again \(x = 7\) and \(x = -9\).
Therefore, the equations \(x^2 - 2x - 63 = 0\) and \(x^2 = -2x + 63\) have the same solutions as \((x−7)(x+9)=0\).
However, since you asked for one option: The correct response is: \(x^2 - 2x - 63 = 0\)