Question 481249
You were correct in your approach...
{{{x^3-5x = 0}}}
{{{x(x^2-5) = 0}}}
set each term (the "x" and the "x^2-5" to zero to get:
x^2-5 = 0
x^2 = 5
x = +-sqrt(5)
.
and, of course
x = 0
.
so, the solution set is:
x = {-sqrt(5), 0, sqrt(5)}