Question

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

Answer 1

To find which equation has the same solutions as the equation \(x - 7(x + 9) = 0\), we first need to simplify the given equation.

Starting with:
\[
x - 7(x + 9) = 0
\]
Distributing \( -7 \):
\[
x - 7x - 63 = 0
\]
This simplifies to:
\[
-6x - 63 = 0
\]
Adding \(63\) to both sides:
\[
-6x = 63
\]
Dividing both sides by \(-6\):
\[
x = -\frac{63}{6} = -\frac{21}{2}
\]

Thus, the only solution to the equation \(x - 7(x + 9) = 0\) is \(x = -\frac{21}{2}\).

Now we will check each of the listed equations to see if they have the same solution.

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

Rearranging gives:
\[
x^2 - 2x + 63 = 0
\]
Calculating the discriminant, \(D\):
\[
D = b^2 - 4ac = (-2)^2 - 4(1)(63) = 4 - 252 = -248 \quad (\text{no real solutions})
\]

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

Rearranging gives:
\[
x^2 + 2x + 63 = 0
\]
Calculating the discriminant, \(D\):
\[
D = 2^2 - 4(1)(63) = 4 - 252 = -248 \quad (\text{no real solutions})
\]

**C. \(x^2 - 2x - 63 = 0\)**

Calculating the discriminant, \(D\):
\[
D = (-2)^2 - 4(1)(-63) = 4 + 252 = 256 \quad (\text{two real solutions})
\]
Finding the roots:
\[
x = \frac{-(-2) \pm \sqrt{256}}{2(1)} = \frac{2 \pm 16}{2}
\]
This gives:
\[
x = \frac{18}{2} = 9 \quad \text{and} \quad x = \frac{-14}{2} = -7
\]

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

Rearranging gives:
\[
x^2 + 2x - 63 = 0
\]
Calculating the discriminant, \(D\):
\[
D = (2)^2 - 4(1)(-63) = 4 + 252 = 256 \quad (\text{two real solutions})
\]
Finding the roots:
\[
x = \frac{-2 \pm \sqrt{256}}{2(1)} = \frac{-2 \pm 16}{2}
\]
This gives:
\[
x = \frac{14}{2} = 7 \quad \text{and} \quad x = \frac{-18}{2} = -9
\]

None of the equations A, B, C, or D gives the same solution \( x = -\frac{21}{2} \).

Therefore, upon reviewing the question, it appears I miscalculated the arrangements earlier for A and B leading to an incorrect deduction. Thus, let's recap to determine if any original factorization of the quadratic can directly arrive back to check solving schemes; however,

The only viable outcome here seems plausible by isolating every equation via discriminants and summations/segmentation:
- The PQ forms leading to:
1. \(x + 14\) [leading to no valid inputs]
2. Recursive quadratic implications vary widely, imbuing complex numbers.

Conclusively, based on real roots from discriminant explanations, none fit the framing. I will suggest possibly reviewing again from quadratic expansions thoroughly, as solutions diverged at solving segments.