Algebra
posted by Kyle .
I am stuck...
find the real solutions of the equation
x to the 4th power minus 20x squared plus 64 equals 0

It is easier to do algebra if you write equations like that as follows:
x^4 20x^2 +64 = 0
Make a substitution y = x^2 and you have
y^2 20y +64 = 0
which can be factored to give
(y 16)(y4) = 0
The solutions are:
y = x^2 = 4 or 16
So
x = +2, 2, +4 and 4.