Question 233189: Hello.
Can you help me figure out what I am doing wrong?
I need to find the Zeros of the following function.
I am getting 0(multiplicity 3), -3, 3.
Thank you in advance.
William
Found 2 solutions by edjones, Theo:
Answer by edjones(8007)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! y = 3x^5 - 9x^3
Set the equation equal to 0 to get:
3x^5 - 9x^3 = 0
factor out x^3 to get:
x^3 * (3x^2 - 9) = 0
This means that x^3 = 0 or (3x^2-9) = 0 or both.
If x^3 = 0, then x = root(3,0) = 0
If (3x^2-9) = 0, then:
Divide by 3 to get:
x^2 - 3 = 0
Add 9 to both sides to get:
x^2 = 3
Take square root of both sides to get:
x = +/- sqrt(3)
Your answers should be that x = 0 or x = sqrt(3) or x = -sqrt(3)
Plug these values into the original equation to see if the equation is true.
With x = 0, 3x^5-9x^3 becomes 0 so this part is true.
With x = sqrt(3), 3x^5-9x^3 becomes 3*(sqrt(3))^5 - 9*(sqrt(3))^3
This becomes 3*3*3*sqrt(3) - 9*3*sqrt(3) = 27*sqrt(3) - 27*sqrt(3) = 0 so this part is true.
With x = -sqrt(3), 3x^5 - 9x^3 becomes 3 * (-sqrt(3))^5 - 3*(-sqrt(3))^3
This becomes 3*3*3*(-sqrt(3)) - 9*3*(-sqrt(3)) = -27*(sqrt(3)) + (27*(sqrt(3) = 0 so this part is true.
Your answers are:
0,-sqrt(3),sqrt(3)
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