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:
- \(2x - 1 = 0\)
- \(x^2 + 3x - 4 = 0\)
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:
- \(x = \frac{-3 + 5}{2} = \frac{2}{2} = 1\)
- \(x = \frac{-3 - 5}{2} = \frac{-8}{2} = -4\)
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., \(\frac{1}{2}, -4, 1\)).
1 answer