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

To solve the equation \( x^2(2x - 1) + 3x(2x - 1) - 4(2x - 1) = 0 \), we can factor out the common term, which is \( (2x - 1) \).

We can rewrite the equation as:

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

This implies two cases:

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

**Solving the first case:**
\[
2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}
\]

**Solving the second case:**
For \( x^2 + 3x - 4 = 0 \), we can factor it:

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

Thus, we have two more solutions:

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

Combining all solutions, we have:

1. \( x = \frac{1}{2} \)
2. \( x = -4 \)
3. \( x = 1 \)

The complete set of solutions is:

\[
\left\{ \frac{1}{2}, -4, 1 \right\}
\]

Now we can check the answer choices:

A: \( 12, 4, -1\frac{1}{2}, 4, -1 \)

B: \( 12, -4, 1\frac{1}{2}, -4, 1 \)

C: \( -4, 1\)

D: \( 12\frac{1}{2} \)

None of the sets provided in the options seems to perfectly match the correct solutions of \( \frac{1}{2}, -4, 1 \).

However, the closest option in terms of included values appears to be option C, which at least contains two of the correct values, \( -4 \) and \( 1 \).

So, none of the answer choices fully match the complete solution set, but option C contains part of it. If forced to choose from the options given, I would choose:

C: \( -4, 1 \) as they are part of the complete set of solutions.