Question

Janet is trying to solve an equation and gets stuck at x^2 ( 2x - 1) + 3x (2x - 1) - 4 (2x - 1) = 0
. Which of the following is a complete set of solutions to Janet’s equation?

A
12, 4, −1\frac{1}{2},\ 4,\ -1
2
1
​
, 4, −1

B
12, −4, 1\frac{1}{2},\ -4,\ 1
2
1
​
, −4, 1

C
−4, 1-4,\ 1−4, 1

D
12\frac{1}{2}
2
1
​

Answer 1

Let's first simplify Janet's equation step by step:

The equation given is:

\[ x^2(2x - 1) + 3x(2x - 1) - 4(2x - 1) = 0. \]

We can factor out \( (2x - 1) \) from each term:

\[
(2x - 1)(x^2 + 3x - 4) = 0.
\]

This gives us two equations to solve:

1. \( 2x - 1 = 0 \)
2. \( x^2 + 3x - 4 = 0 \)

### Solving the first equation:
For \( 2x - 1 = 0 \):
\[
2x = 1 \\
x = \frac{1}{2}.
\]

### Solving the second equation:
For \( x^2 + 3x - 4 = 0 \):
We can factor this quadratic:

\[
(x + 4)(x - 1) = 0.
\]

Setting each factor to zero gives us:

1. \( x + 4 = 0 \) which gives \( x = -4 \)
2. \( x - 1 = 0 \) which gives \( x = 1 \)

### Complete set of solutions:
Now we combine the solutions from both equations:

- From \( 2x - 1 = 0\): \( x = \frac{1}{2} \)
- From \( x^2 + 3x - 4 = 0\): \( x = -4, 1 \)

Thus, the complete set of solutions is:

\[
x = \frac{1}{2}, -4, 1.
\]

### Analyzing the Answer Choices:
Let's analyze the answer choices provided:

- A: \( 12, 4, -1\frac{1}{2}, 4, -1 \) (Does not match)
- B: \( 12, -4, 1\frac{1}{2}, -4, 1 \) (Does not match)
- C: \( -4, 1 \) (Includes part of the answer)
- D: \( 12\frac{1}{2} \) (Does not match)

Thus, the correct complete set of solutions is not explicitly listed among the options. The final set of solutions we computed is:

\[
x = \frac{1}{2}, -4, 1.
\]