8x^3+4x^2-4x solution sets
solution sets? Is there an equals sign anywhere?
If all this is equal to zero..
4x(2x^2+x-1)=4x(2x-1)(x+1)=0
x=0, x=1/2, x=-1
check those
thank you!
To find the solution sets for the equation 8x^3 + 4x^2 - 4x, we need to set the equation equal to zero and solve for x.
Step 1: Set the equation equal to zero:
8x^3 + 4x^2 - 4x = 0
Step 2: Factor out common terms, if possible:
We can see that there is a common factor of 4x, so we can factor that out:
4x(2x^2 + x - 1) = 0
Step 3: Solve for x by setting each factor equal to zero:
4x = 0 or 2x^2 + x - 1 = 0
Solving the first equation, we get:
x = 0
To solve the second equation, we can use the quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / (2a)
In this case, a = 2, b = 1, and c = -1. Plugging these values into the quadratic formula, we get:
x = [-1 ± sqrt(1 - 4(2)(-1))] / (2 * 2)
x = [-1 ± sqrt(1 + 8)] / 4
x = [-1 ± sqrt(9)] / 4
x = [-1 ± 3] / 4
So, the solutions for the second equation are:
x = (-1 + 3) / 4 = 1/2
x = (-1 - 3) / 4 = -1
Therefore, the solution set for the equation 8x^3 + 4x^2 - 4x = 0 is:
{x = 0, x = 1/2, x = -1}