Question
5. Which of the following quadratic equations is not solvable by grouping?(1 point)
Responses
2x2+14x+12=0
2
2
+
14
+
12
=
0
2 x squared plus 14 x plus 12 is equal to 0
x2−12x+35=0
2
−
12
+
35
=
0
x squared minus 12 x plus 35 is equal to 0
x2−2x+1=0
2
−
2
+
1
=
0
x squared minus 2 x plus 1 is equal to 0
2x2−2x−10=0
Responses
2x2+14x+12=0
2
2
+
14
+
12
=
0
2 x squared plus 14 x plus 12 is equal to 0
x2−12x+35=0
2
−
12
+
35
=
0
x squared minus 12 x plus 35 is equal to 0
x2−2x+1=0
2
−
2
+
1
=
0
x squared minus 2 x plus 1 is equal to 0
2x2−2x−10=0
Answers
Answered by
GPT-4o mini
To determine which of the given quadratic equations is not solvable by grouping, we need to check if there are pairs of terms in each equation that can be factored or grouped effectively.
1. **Equation**: \(2x^2 + 14x + 12 = 0\)
Factoring out 2 gives:
\(2(x^2 + 7x + 6) = 0\)
Further factoring \(x^2 + 7x + 6 = (x + 1)(x + 6)\).
2. **Equation**: \(x^2 - 12x + 35 = 0\)
This can be factored as:
\((x - 5)(x - 7) = 0\).
3. **Equation**: \(x^2 - 2x + 1 = 0\)
This can be factored as:
\((x - 1)(x - 1) = 0\) or \((x - 1)^2 = 0\).
4. **Equation**: \(2x^2 - 2x - 10 = 0\)
Factoring out 2 gives:
\(2(x^2 - x - 5) = 0\).
The quadratic \(x^2 - x - 5\) does not factor nicely with integer pairs, and results in irrational solutions.
From this analysis:
- The first three equations can be factored or grouped satisfactorily.
- The fourth equation, \(2x^2 - 2x - 10 = 0\), does not factor nicely and is best solved using the quadratic formula rather than grouping.
Therefore, the equation that is **not solvable by grouping** is:
**2x^2 - 2x - 10 = 0**.
1. **Equation**: \(2x^2 + 14x + 12 = 0\)
Factoring out 2 gives:
\(2(x^2 + 7x + 6) = 0\)
Further factoring \(x^2 + 7x + 6 = (x + 1)(x + 6)\).
2. **Equation**: \(x^2 - 12x + 35 = 0\)
This can be factored as:
\((x - 5)(x - 7) = 0\).
3. **Equation**: \(x^2 - 2x + 1 = 0\)
This can be factored as:
\((x - 1)(x - 1) = 0\) or \((x - 1)^2 = 0\).
4. **Equation**: \(2x^2 - 2x - 10 = 0\)
Factoring out 2 gives:
\(2(x^2 - x - 5) = 0\).
The quadratic \(x^2 - x - 5\) does not factor nicely with integer pairs, and results in irrational solutions.
From this analysis:
- The first three equations can be factored or grouped satisfactorily.
- The fourth equation, \(2x^2 - 2x - 10 = 0\), does not factor nicely and is best solved using the quadratic formula rather than grouping.
Therefore, the equation that is **not solvable by grouping** is:
**2x^2 - 2x - 10 = 0**.