Asked by LL
Which of the fallowing equations has exactly the same solution as (x-7)(x+9)=0?
A. x^2=2x-63
B. x^2=-2x+63
C. x^2+2x=-63
D. x^2-2x-63=0
A. x^2=2x-63
B. x^2=-2x+63
C. x^2+2x=-63
D. x^2-2x-63=0
Answers
Answered by
GPT-4o mini
To determine which of the given equations has the same solution as \((x-7)(x+9)=0\), we first need to solve the original equation.
The solutions to \((x-7)(x+9)=0\) can be found by setting each factor equal to zero:
1. \(x - 7 = 0 \Rightarrow x = 7\)
2. \(x + 9 = 0 \Rightarrow x = -9\)
So, the solutions are \(x = 7\) and \(x = -9\).
Next, we will check each of the options to see if they also yield these same solutions.
**Option A:** \(x^2 = 2x - 63\)
Rearranging gives:
\[ x^2 - 2x + 63 = 0 \]
Using the quadratic formula to find the solutions:
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 63}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 252}}{2} = \frac{2 \pm \sqrt{-248}}{2} \]
Since the discriminant is negative, this does not have real solutions.
**Option B:** \(x^2 = -2x + 63\)
Rearranging gives:
\[ x^2 + 2x - 63 = 0 \]
Using the quadratic formula:
\[ x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-63)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 252}}{2} = \frac{-2 \pm \sqrt{256}}{2} = \frac{-2 \pm 16}{2} \]
Calculating:
1. \(x = \frac{14}{2} = 7\)
2. \(x = \frac{-18}{2} = -9\)
This option has the same solutions \(x = 7\) and \(x = -9\).
**Option C:** \(x^2 + 2x = -63\)
Rearranging gives:
\[ x^2 + 2x + 63 = 0 \]
Using the quadratic formula:
\[ x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot 63}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 - 252}}{2} = \frac{-2 \pm \sqrt{-248}}{2} \]
This again yields no real solutions.
**Option D:** \(x^2 - 2x - 63 = 0\)
Using the quadratic formula:
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-63)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 252}}{2} = \frac{2 \pm \sqrt{256}}{2} = \frac{2 \pm 16}{2} \]
Calculating:
1. \(x = \frac{18}{2} = 9\)
2. \(x = \frac{-14}{2} = -7\)
These solutions are \(x = 9\) and \(x = -7\), which do not match.
In conclusion, the equation that has exactly the same solutions as \((x-7)(x+9)=0\) is option **B**: \(x^2 = -2x + 63\).
The solutions to \((x-7)(x+9)=0\) can be found by setting each factor equal to zero:
1. \(x - 7 = 0 \Rightarrow x = 7\)
2. \(x + 9 = 0 \Rightarrow x = -9\)
So, the solutions are \(x = 7\) and \(x = -9\).
Next, we will check each of the options to see if they also yield these same solutions.
**Option A:** \(x^2 = 2x - 63\)
Rearranging gives:
\[ x^2 - 2x + 63 = 0 \]
Using the quadratic formula to find the solutions:
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 63}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 252}}{2} = \frac{2 \pm \sqrt{-248}}{2} \]
Since the discriminant is negative, this does not have real solutions.
**Option B:** \(x^2 = -2x + 63\)
Rearranging gives:
\[ x^2 + 2x - 63 = 0 \]
Using the quadratic formula:
\[ x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-63)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 252}}{2} = \frac{-2 \pm \sqrt{256}}{2} = \frac{-2 \pm 16}{2} \]
Calculating:
1. \(x = \frac{14}{2} = 7\)
2. \(x = \frac{-18}{2} = -9\)
This option has the same solutions \(x = 7\) and \(x = -9\).
**Option C:** \(x^2 + 2x = -63\)
Rearranging gives:
\[ x^2 + 2x + 63 = 0 \]
Using the quadratic formula:
\[ x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot 63}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 - 252}}{2} = \frac{-2 \pm \sqrt{-248}}{2} \]
This again yields no real solutions.
**Option D:** \(x^2 - 2x - 63 = 0\)
Using the quadratic formula:
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-63)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 252}}{2} = \frac{2 \pm \sqrt{256}}{2} = \frac{2 \pm 16}{2} \]
Calculating:
1. \(x = \frac{18}{2} = 9\)
2. \(x = \frac{-14}{2} = -7\)
These solutions are \(x = 9\) and \(x = -7\), which do not match.
In conclusion, the equation that has exactly the same solutions as \((x-7)(x+9)=0\) is option **B**: \(x^2 = -2x + 63\).
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