Question 695453
find the real number solutions
:
8x^4 - x = 0
Factor out x
x(8x^3 - 1) = 0
this is difference of cubes, and can be factored
x(2x-1)(4x^2 + x + 1) 
Two real solutions; (4x^2+x+1) will not yield a real solution)
x = 0
and
2x = 1
x = .5
:
and
x^4 + 7x^2 - 18 = 0
Factors to
(x^2 + 9)(x^2 - 2) = 0
Two real solutions, (x^2+9) will not yield real solutions
x^2 = 2
x = +{{{sqrt(2)}}}
and
x = -{{{sqrt(2)}}}