To find which equation has exactly the same solutions as \((x−7)(x+9)=0\), we first need to solve the original equation.
The equation \((x−7)(x+9)=0\) is satisfied when either factor is equal to zero:
- \(x - 7 = 0 \Rightarrow x = 7\)
- \(x + 9 = 0 \Rightarrow x = -9\)
So, the solutions are \(x = 7\) and \(x = -9\).
Now, let's check each option to see which one has the same solutions.
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Option 1: \(x^2 = 2x - 63\)
Rearranging gives us:
\(x^2 - 2x + 63 = 0\)
We can use the quadratic formula to find the roots: \[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 63}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 252}}{2}\]
Since the discriminant \(4 - 252 < 0\), there are no real solutions. -
Option 2: \(x^2 - 2x - 63 = 0\)
We can factor this:
\((x - 9)(x + 7) = 0\)
This gives us the solutions \(x = 9\) and \(x = -7\), which do not match. -
Option 3: \(x^2 = -2x + 63\)
Rearranging gives us:
\(x^2 + 2x - 63 = 0\)
We can factor this:
\((x - 7)(x + 9) = 0\)
This gives us the solutions \(x = 7\) and \(x = -9\), which match! -
Option 4: \(x^2 + 2x = -63\)
Rearranging gives us:
\(x^2 + 2x + 63 = 0\)
The discriminant is \(2^2 - 4 \cdot 1 \cdot 63 = 4 - 252 < 0\), so there are no real solutions.
Based on the above analyses, the equation with exactly the same solutions as \((x−7)(x+9)=0\) is:
Option 3: \(x^2 + 2x - 63 = 0\).