Which of the following quadratic equations is not solvable by grouping?(1 point)

Question

Responses

2x2+14x+12=0
2 x squared plus 14 x plus 12 equals 0

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

Answer 1

To determine which of the given quadratic equations is not solvable by grouping, we first check if the equations can be factored by grouping. Grouping typically involves rearranging and extracting common factors in pairs of terms.

1. Equation: \(2x^2 + 14x + 12 = 0\)

   • This can be factored as follows:
   • Factor out a 2: \(2(x^2 + 7x + 6) = 0\)
   • Further factoring gives \(2(x + 6)(x + 1) = 0\).
   • This is solvable by grouping.

2. Equation: \(x^2 - 12x + 35 = 0\)

   • This can be factored as:
   • \((x - 5)(x - 7) = 0\).
   • This is also solvable by grouping.

3. Equation: \(2x^2 - 2x - 10 = 0\)

   • Factor out a 2: \(2(x^2 - x - 5) = 0\).
   • The quadratic \(x^2 - x - 5\) doesn't factor nicely into integers, hence it's not easily solvable by grouping.

4. Equation: \(x^2 - 2x + 1 = 0\)

   • This can be factored as \((x - 1)(x - 1) = (x - 1)^2 = 0\).
   • This is also solvable by grouping.

Based on this analysis, the equation that is not solvable by grouping is:

\(2x^2 - 2x - 10 = 0\).

Answer 2

Are you sure?

Answer 3

Let's take another look at each of the quadratic equations to confirm which one is not easily factored by grouping.

1. Equation: \(2x^2 + 14x + 12 = 0\)

   • Factoring out a 2 gives \(2(x^2 + 7x + 6) = 0\).
   • The quadratic \(x^2 + 7x + 6\) factors to \((x + 6)(x + 1)\).
   • So, this equation is solvable by grouping.

2. Equation: \(x^2 - 12x + 35 = 0\)

   • This factors to \((x - 5)(x - 7) = 0\).
   • Therefore, this equation is also solvable by grouping.

3. Equation: \(2x^2 - 2x - 10 = 0\)

   • Factoring out a 2 gives \(2(x^2 - x - 5) = 0\).
   • The quadratic \(x^2 - x - 5\) does not factor into rational numbers, making it more challenging to solve by grouping.
   • We can use the quadratic formula, but it doesn’t factor nicely.

4. Equation: \(x^2 - 2x + 1 = 0\)

   • This is a perfect square: \((x - 1)^2 = 0\).
   • It is easily solvable by grouping.

After revisiting the equations, I can reaffirm that the equation \(2x^2 - 2x - 10 = 0\) is indeed the one that is not easily solvable by grouping, as it results in a quadratic that does not factor into integers. Thank you for your patience!