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which of the following quadratic equations is NOT solvable by grouping 2x^2 - 2x - 10 = 0 x^2 - 12x + 35 = 0 2x^2 + 14x + 12 =...Question
which of the following quadratic equations is NOT solvable by grouping
2x^2 - 2x - 10 = 0
x^2 - 12x + 35 = 0
2x^2 + 14x + 12 = 0
x^2 - 2x + 1 = 0
2x^2 - 2x - 10 = 0
x^2 - 12x + 35 = 0
2x^2 + 14x + 12 = 0
x^2 - 2x + 1 = 0
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GPT-4o mini
To determine which quadratic equation is NOT solvable by grouping, we first need to consider the method of factoring by grouping. This method typically works when the polynomial can be rearranged into two pairs of terms that can be factored separately.
Let's analyze each quadratic equation:
1. **\(2x^2 - 2x - 10 = 0\)**
- This can be factored as \(2(x^2 - x - 5) = 0\). The quadratic \(x^2 - x - 5\) does not factor nicely, but it's still possible to find roots using the quadratic formula if needed.
2. **\(x^2 - 12x + 35 = 0\)**
- This can be factored as \((x - 7)(x - 5) = 0\). It factors nicely.
3. **\(2x^2 + 14x + 12 = 0\)**
- This can be factored by first factoring out 2: \(2(x^2 + 7x + 6) = 0\). The quadratic \(x^2 + 7x + 6\) can then be factored as \((x + 6)(x + 1) = 0\). This one factors nicely.
4. **\(x^2 - 2x + 1 = 0\)**
- This can be factored as \((x - 1)(x - 1) = 0\) or \((x - 1)^2 = 0\). This also factors nicely.
Out of these equations, the first one, **\(2x^2 - 2x - 10 = 0\)**, is not easily solvable by grouping because the quadratic \(x^2 - x - 5\) does not factor into rational numbers, making it less suited for factoring by grouping compared to the others which factor straightforwardly.
So, the quadratic equation that is NOT solvable by grouping is:
**\(2x^2 - 2x - 10 = 0\)**.
Let's analyze each quadratic equation:
1. **\(2x^2 - 2x - 10 = 0\)**
- This can be factored as \(2(x^2 - x - 5) = 0\). The quadratic \(x^2 - x - 5\) does not factor nicely, but it's still possible to find roots using the quadratic formula if needed.
2. **\(x^2 - 12x + 35 = 0\)**
- This can be factored as \((x - 7)(x - 5) = 0\). It factors nicely.
3. **\(2x^2 + 14x + 12 = 0\)**
- This can be factored by first factoring out 2: \(2(x^2 + 7x + 6) = 0\). The quadratic \(x^2 + 7x + 6\) can then be factored as \((x + 6)(x + 1) = 0\). This one factors nicely.
4. **\(x^2 - 2x + 1 = 0\)**
- This can be factored as \((x - 1)(x - 1) = 0\) or \((x - 1)^2 = 0\). This also factors nicely.
Out of these equations, the first one, **\(2x^2 - 2x - 10 = 0\)**, is not easily solvable by grouping because the quadratic \(x^2 - x - 5\) does not factor into rational numbers, making it less suited for factoring by grouping compared to the others which factor straightforwardly.
So, the quadratic equation that is NOT solvable by grouping is:
**\(2x^2 - 2x - 10 = 0\)**.
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