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Which of the following quadratic equations is not solvable by grouping?(1 point) Responses x2−12x+35=0 x squared minus 12 x plu...Question
Which of the following quadratic equations is not solvable by grouping?(1 point)
Responses
x2−12x+35=0
x squared minus 12 x plus 35 equals 0
2x2−2x−10=0
2 x squared minus 2 x minus 10 equals 0
x2−2x+1=0
x squared minus 2 x plus 1 equals 0
2x2+14x+12=0
Responses
x2−12x+35=0
x squared minus 12 x plus 35 equals 0
2x2−2x−10=0
2 x squared minus 2 x minus 10 equals 0
x2−2x+1=0
x squared minus 2 x plus 1 equals 0
2x2+14x+12=0
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To determine which quadratic equation is not solvable by grouping, we can look at each given option and see if they can be factored by grouping.
1. **\(x^2 - 12x + 35 = 0\)**:
- This factors as \((x - 5)(x - 7) = 0\).
- It can be solved, so it is solvable by grouping.
2. **\(2x^2 - 2x - 10 = 0\)**:
- First, factor out 2: \(2(x^2 - x - 5) = 0\).
- The quadratic \(x^2 - x - 5\) does not factor nicely. However, we can use the quadratic formula. Factoring via grouping will not work here conveniently.
- Additionally, it can be rewritten as \(2(x - 5)(x + 1)\) using other methods.
3. **\(x^2 - 2x + 1 = 0\)**:
- This factors as \((x - 1)(x - 1) = 0\) or \((x - 1)^2 = 0\).
- It can be solved, so it is solvable by grouping.
4. **\(2x^2 + 14x + 12 = 0\)**:
- First, factor out 2: \(2(x^2 + 7x + 6) = 0\).
- The quadratic \(x^2 + 7x + 6\) can be factored as \((x + 1)(x + 6) = 0\).
- It can be solved, so it is solvable by grouping.
After analyzing these equations, the one that is not likely to be effectively solvable by grouping (as it does not factor easily into a form that reveals straightforward pairs) is:
**2x^2−2x−10=0**
Thus, the final answer is: **2x² - 2x - 10 = 0**.
1. **\(x^2 - 12x + 35 = 0\)**:
- This factors as \((x - 5)(x - 7) = 0\).
- It can be solved, so it is solvable by grouping.
2. **\(2x^2 - 2x - 10 = 0\)**:
- First, factor out 2: \(2(x^2 - x - 5) = 0\).
- The quadratic \(x^2 - x - 5\) does not factor nicely. However, we can use the quadratic formula. Factoring via grouping will not work here conveniently.
- Additionally, it can be rewritten as \(2(x - 5)(x + 1)\) using other methods.
3. **\(x^2 - 2x + 1 = 0\)**:
- This factors as \((x - 1)(x - 1) = 0\) or \((x - 1)^2 = 0\).
- It can be solved, so it is solvable by grouping.
4. **\(2x^2 + 14x + 12 = 0\)**:
- First, factor out 2: \(2(x^2 + 7x + 6) = 0\).
- The quadratic \(x^2 + 7x + 6\) can be factored as \((x + 1)(x + 6) = 0\).
- It can be solved, so it is solvable by grouping.
After analyzing these equations, the one that is not likely to be effectively solvable by grouping (as it does not factor easily into a form that reveals straightforward pairs) is:
**2x^2−2x−10=0**
Thus, the final answer is: **2x² - 2x - 10 = 0**.
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