Question

Which of the following equations has exactly the same solutions as (x - 7) (x + 9) = 0
A. x^2 - 2x - 63 = 0
B. x^2 + 2x = -63
C. x^2 = 2x - 63
D. x^2 = -2x + 63

Answer 1

To find the equation that has exactly the same solutions as \((x - 7)(x + 9) = 0\), we first solve for the values of \(x\) that satisfy this equation.

Setting each factor equal to zero:

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

So the solutions to the equation \((x - 7)(x + 9) = 0\) are \(x = 7\) and \(x = -9\).

Now we will check each option to see if it produces the same solutions.

**A. \(x^2 - 2x - 63 = 0\)**
To find the solutions, we can factor this or use the quadratic formula. Factoring, we look for two numbers that multiply to \(-63\) and add to \(-2\). The factors are \(7\) and \(-9\):
\[
(x - 9)(x + 7) = 0
\]
Setting each factor to zero:
1. \(x - 9 = 0 \Rightarrow x = 9\)
2. \(x + 7 = 0 \Rightarrow x = -7\)

This does not match our solutions.

**B. \(x^2 + 2x = -63\)**
Rearranging gives:
\[
x^2 + 2x + 63 = 0
\]
Using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4(1)(63)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 252}}{2} = \frac{-2 \pm \sqrt{-248}}{2}
\]
Since we have a negative discriminant, this equation has no real solutions.

**C. \(x^2 = 2x - 63\)**
Rearranging gives:
\[
x^2 - 2x + 63 = 0
\]
Using the quadratic formula here as well:
\[
x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(63)}}{2(1)} = \frac{2 \pm \sqrt{4 - 252}}{2} = \frac{2 \pm \sqrt{-248}}{2}
\]
Again, we have a negative discriminant, so no real solutions.

**D. \(x^2 = -2x + 63\)**
Rearranging gives:
\[
x^2 + 2x - 63 = 0
\]
Factoring this:
\[
(x + 9)(x - 7) = 0
\]
Setting the factors to zero:
1. \(x + 9 = 0 \Rightarrow x = -9\)
2. \(x - 7 = 0 \Rightarrow x = 7\)

This matches the solutions from the original equation.

Hence, the correct answer is **D. \(x^2 = -2x + 63\)**.