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:
- \(x + 14\) [leading to no valid inputs]
- 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.