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?
To solve the equation, we can factor out the common factor of (2x - 1):
(2x - 1)(x^2 + 3x - 4) = 0
Now, we can set each factor equal to zero and solve for x:
2x - 1 = 0
x = 1/2
x^2 + 3x - 4 = 0
(x + 4)(x - 1) = 0
x = -4, 1
Therefore, the complete set of solutions to Janet's equation is x = -4, 1/2, 1.
To solve the equation x^2 (2x - 1) + 3x (2x - 1) - 4 (2x - 1) = 0, we can factor out the common term (2x - 1):
(2x - 1) (x^2 + 3x - 4) = 0
Now we have a product of two terms equal to zero. This means that either (2x - 1) = 0 or (x^2 + 3x - 4) = 0.
Solving for (2x - 1) = 0, we get:
2x - 1 = 0
2x = 1
x = 1/2
Solving for (x^2 + 3x - 4) = 0, we can factor the quadratic:
(x + 4) (x - 1) = 0
Setting each factor equal to zero, we have two equations:
x + 4 = 0
x = -4
x - 1 = 0
x = 1
Therefore, Janet's equation has three solutions: x = 1/2, x = -4, and x = 1.
To solve the given equation, we can use a technique called factoring. Let's go step by step:
Step 1: Let's observe that the equation can be rewritten as follows:
(2x - 1) (x^2 + 3x - 4) = 0
Step 2: Now we have a product equal to zero. According to the zero product property, if a product of two or more factors equals zero, then at least one of the factors must be zero.
So, we can write two separate equations:
1. 2x - 1 = 0
2. x^2 + 3x - 4 = 0
Step 3: Solving the equations separately:
1. To solve 2x - 1 = 0, we add 1 to both sides and then divide by 2:
2x - 1 + 1 = 0 + 1
2x = 1
x = 1/2
2. To solve x^2 + 3x - 4 = 0, we can factor this quadratic equation:
(x + 4)(x - 1) = 0
Using the zero product property again, we have two possibilities:
a) x + 4 = 0, which gives x = -4
b) x - 1 = 0, which gives x = 1
So, the solutions to the equation are: x = 1/2, x = -4, and x = 1.
In summary, the complete set of solutions to Janet's equation is {1/2, -4, 1}.