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)(y-4) = 0
The solutions are:
y = x^2 = 4 or 16
So
x = +2, -2, +4 and -4.

To find the real solutions of the equation, you can use the technique of factoring. However, in this case, the equation is a quadratic in terms of x^2. We can rewrite it as follows:

x^4 - 20x^2 + 64 = 0

Let's simplify the equation by substituting a variable. Let's say, let y = x^2. Now we can rewrite the equation in terms of y:

y^2 - 20y + 64 = 0

This is now a simple quadratic equation. To solve it, we can use factoring or the quadratic formula. In this case, let's try factoring.

We need to find two numbers that multiply to 64 and add up to -20. The numbers -4 and -16 satisfy these conditions:

(y - 4)(y - 16) = 0

To find the values of y (or x^2), we can set each factor equal to zero:

y - 4 = 0
y - 16 = 0

Solving these equations, we find:

y = 4
y = 16

Since y = x^2, we can solve for x:

For y = 4:
x^2 = 4
Taking the square root of both sides, we get:
x = ±2

For y = 16:
x^2 = 16
Taking the square root of both sides, we get:
x = ±4

Therefore, the real solutions of the equation x^4 - 20x^2 + 64 = 0 are x = -4, -2, 2, and 4.