Question

Janet is trying to solve an equation and gets stuck at x^2(2x-1)+3x(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 \( x^2(2x-1) + 3x(2x-1) = 0 \), we can factor this equation.

First, we notice that \(2x - 1\) is a common factor:

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

Now, we can set each factor equal to zero:

1. \(2x - 1 = 0\)
- Solving this gives:
\[
2x = 1 \implies x = \frac{1}{2}
\]

2. \(x^2 + 3x = 0\)
- Factoring out \(x\) gives:
\[
x(x + 3) = 0
\]
- Setting each factor to zero gives:
- \(x = 0\)
- \(x + 3 = 0 \implies x = -3\)

So, the complete set of solutions to the equation is:
\[
x = \frac{1}{2}, 0, -3
\]

Now, we check the options:

- A: \(12, 4, -1\frac{1}{2}, 4, -1\) - This does not match
- B: \(12, -4, 1\frac{1}{2}, -4, 1\) - This does not match
- C: \(-4, 1\) - This does not match
- D: \(12\frac{1}{2}\) - This does not match

None of the options A, B, C, or D are correct based on the solutions we've found. The correct solutions are \(x = \frac{1}{2}, 0, -3\).