Question

Janet is trying to solve an equation and gets stuck at x2(2x−1)+3x(2x−1)−4(2x−1)=0
. Which of the following is a complete set of solutions to Janet’s equation?(1 point)
Responses

12
, −4
, 1
Start Fraction 1 over 2 End Fraction , negative 4 , 1

−4
, 1
negative 4 , 1

12
, 4, −1
Start Fraction 1 over 2 End Fraction , 4, negative 1

12

Answer 1

To solve the equation provided by Janet, we start with:

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

First, we can factor out the common term \((2x - 1)\):

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

Now we have a product equal to zero, which means we can set each factor to zero:

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

### Solving the first factor:

From \(2x - 1 = 0\):

\[
2x = 1 \quad \Rightarrow \quad x = \frac{1}{2}.
\]

### Solving the second factor:

Now, we solve the quadratic equation \(x^2 + 3x - 4 = 0\) using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\]

where \(a = 1\), \(b = 3\), and \(c = -4\).

Calculating the discriminant:

\[
b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot (-4) = 9 + 16 = 25.
\]

Now applying the quadratic formula:

\[
x = \frac{-3 \pm \sqrt{25}}{2 \cdot 1} = \frac{-3 \pm 5}{2}.
\]

This gives us two solutions:

1. \(x = \frac{-3 + 5}{2} = \frac{2}{2} = 1\)
2. \(x = \frac{-3 - 5}{2} = \frac{-8}{2} = -4\)

### Complete Set of Solutions:

Combining all the solutions, we get:

- From the first factor: \(x = \frac{1}{2}\)
- From the second factor: \(x = 1\) and \(x = -4\)

Thus, the complete set of solutions is:

\(\frac{1}{2}, 1, -4\).

Looking at the provided options, the correct choice is:

**Start Fraction 1 over 2 End Fraction , negative 4 , 1** (i.e., \(\frac{1}{2}, -4, 1\)).