Question

Which of the following equations has exactly the same solutions as (x−7)(x+9)=0?(1 point)
Responses

x2=−2x+63
x squared equals negative 2 x plus 63

x2=2x−63
x squared equals 2 x minus 63

x2−2x−63=0
x squared minus 2 x minus 63 equals 0

x2+2x=−63

Answer 1

To find the equation that has exactly the same solutions as \((x−7)(x+9)=0\), we can start by solving the original equation.

Setting the factors to zero, we have:
1. \(x - 7 = 0 \Rightarrow x = 7\)
2. \(x + 9 = 0 \Rightarrow x = -9\)

The solutions to \((x−7)(x+9)=0\) are \(x = 7\) and \(x = -9\).

Next, we can check each of the given equations to see if they yield the same solutions:

1. **For \(x^2 = -2x + 63\)**:
\[
x^2 + 2x - 63 = 0 \quad \text{(Rearranging gives us this quadratic equation)}
\]
We can factor this:
\[
(x + 9)(x - 7) = 0
\]
Solutions are \(x = 7\) and \(x = -9\).

2. **For \(x^2 = 2x - 63\)**:
\[
x^2 - 2x + 63 = 0
\]
This does not factor nicely and needs the quadratic formula:
\[
x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 63}}{2 \cdot 1} = \frac{2 \pm \sqrt{-244}}{2}
\]
This has imaginary solutions, not matching \(7\) and \(-9\).

3. **For \(x^2 - 2x - 63 = 0\)**:
Factoring gives:
\[
(x - 9)(x + 7) = 0
\]
Solutions are \(x = 9\) and \(x = -7\), which do not match.

4. **For \(x^2 + 2x = -63\)**:
\[
x^2 + 2x + 63 = 0
\]
This also needs the quadratic formula:
\[
x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot 63}}{2 \cdot 1} = \frac{-2 \pm \sqrt{-244}}{2}
\]
Like before, this does not yield real solutions.

Therefore, the equation that has exactly the same solutions as \((x−7)(x+9)=0\) is:

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