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
Start Fraction 1 over 2 End Fraction
12
, 4, −1
The correct response is: -4, 1, 1/2
To solve the equation x^2(2x−1)+3x(2x−1)−4(2x−1)=0, we need to factor out the common term (2x-1) from each term on the left side of the equation.
Factoring out (2x-1), we get:
(2x-1)(x^2 + 3x - 4) = 0
Now, we can solve for (2x-1) and set it equal to zero or solve for the quadratic equation (x^2 + 3x - 4) = 0.
Setting (2x-1) = 0, we have:
2x = 1
x = 1/2
Now, let's solve the quadratic equation (x^2 + 3x - 4) = 0:
We can factor it as:
(x+4)(x-1) = 0
Setting each factor equal to zero, we have:
x+4 = 0
x = -4
x-1 = 0
x = 1
Therefore, the complete set of solutions to Janet's equation are:
x = 1/2, -4, and 1.
So, the correct option is: Start Fraction 1 over 2 End Fraction , -4 , 1.
To solve the equation x^2(2x-1) + 3x(2x-1) - 4(2x-1) = 0, we need to simplify and rearrange the equation to find the values of x that satisfy it.
First, let's simplify each term in the equation:
x^2(2x-1) can be expanded to 2x^3 - x^2.
3x(2x-1) can be expanded to 6x^2 - 3x.
-4(2x-1) can be expanded to -8x + 4.
Now, we substitute these simplified terms back into the equation:
2x^3 - x^2 + 6x^2 - 3x - 8x + 4 = 0
Combining like terms, we get:
2x^3 + 5x^2 - 11x + 4 = 0
To solve this cubic equation, we can try factoring, but it may not always be possible. In this case, we can use the Rational Root Theorem to determine any possible rational solutions. The theorem states that if a rational number p/q (in simplest form) is a solution to the equation, then p must be a factor of the constant term (in this case, 4), and q must be a factor of the leading coefficient (in this case, 2).
The factors of 4 are ±1, ±2, and ±4.
The factors of 2 are ±1 and ±2.
Therefore, the possible rational solutions are calculated by dividing each factor of the constant term by each factor of the leading coefficient. This gives us the following possibilities for x: ±1/2, ±1, ±2, and ±4.
Now, we can test each of these possibilities by substituting them back into the original equation. It is a bit time-consuming, but we can check if any of the values satisfy the equation:
Let's start with x = 1/2:
Substituting x = 1/2 into the equation:
(1/2)^2(2(1/2)-1) + 3(1/2)(2(1/2)-1) - 4(2(1/2)-1) = 0
(1/4)(1-1) + (3/2)(1-1) - (4)(1-1) = 0
0 + 0 - 0 = 0
Since this equation is true, x = 1/2 is a valid solution.
Testing the other values of ±1/2, ±1, ±2, and ±4 in the same way, we find that the complete set of solutions to Janet’s equation is:
x = 1/2, -4, 1