what are the solution sets to x^5-2x^3+x=0 and how do you get them?
the answers are 1,-1, and 0 but i'm not sure how to explain how i got them.
x^5 - 2x^3 + x = 0
x(x^4 - 2x^2 + 1) = 0
x(x^2 - 1)^2 = 0
so x = 0 or x^2 - 1 = 0 --->(x+1)(x-1) = 0
x = 0 or x =1 or x = -1
To find the solution sets to the equation x^5 - 2x^3 + x = 0, we need to solve for x such that this equation holds true. Here's how you can find the solution sets step-by-step:
Step 1: Factor out common terms
x(x^4 - 2x^2 + 1) = 0
Step 2: Factor the quadratic expression inside the parentheses
x(x^2 - 1)^2 = 0
Step 3: Apply the zero product property
This property states that if a*b = 0, then either a = 0 or b = 0 (or both). Since we have x multiplied by (x^2 - 1)^2, we can set each factor equal to 0 separately.
For x = 0, we have one solution.
For x^2 - 1 = 0, we can solve for x by taking the square root of both sides:
x^2 = 1
x = ±1
However, since the roots (±1) are squared in the equation, they also count twice as solutions. Therefore, we have four additional solutions.
Combining all the solutions, we get the following solution sets:
x = 0, x = 1, x = -1, x = 1, x = -1
Note that x = 1 and x = -1 are repeated in the solution sets since they are squared in the equation.
To summarize, the solution sets to the equation x^5 - 2x^3 + x = 0 are {0, 1, -1}.