Question 233189
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)