To solve Janet's equation \( x^2(2x - 1) + 3x(2x - 1) - 4(2x - 1) = 0 \), we can start by factoring out the common term \( (2x - 1) \):
\[ (2x - 1)(x^2 + 3x - 4) = 0 \]
This gives us two separate equations to solve:
- \( 2x - 1 = 0 \)
- \( x^2 + 3x - 4 = 0 \)
Solving the first equation:
\[ 2x - 1 = 0 \] \[ 2x = 1 \] \[ x = \frac{1}{2} \]
Solving the second equation using the quadratic formula:
The quadratic formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). For the equation \( x^2 + 3x - 4 = 0 \), we have \( a = 1 \), \( b = 3 \), and \( c = -4 \).
Calculating the discriminant:
\[ b^2 - 4ac = 3^2 - 4(1)(-4) = 9 + 16 = 25 \]
Now applying the quadratic formula:
\[ x = \frac{-3 \pm \sqrt{25}}{2(1)} = \frac{-3 \pm 5}{2} \]
Calculating the two possible values:
- \( x = \frac{-3 + 5}{2} = \frac{2}{2} = 1 \)
- \( x = \frac{-3 - 5}{2} = \frac{-8}{2} = -4 \)
Thus, the complete set of solutions is:
\[ x = \frac{1}{2}, x = 1, x = -4 \]
Now, checking the response options to find the complete set of solutions:
The solution set is \( \frac{1}{2}, -4, 1 \).
The correct response from the options provided is:
Start Fraction 1 over 2 End Fraction, −4, 1.