Question 906173
your problem is:


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


solve for x^4 - 20x^2 + 64 = 0 to find out when the graph crosses the x-axis.


this is a quadratic equation in disguise.


let y = x^2 and the equation becomes y^2 - 20y + 64 = 0


factor this equation to get (y-4)*(y-16)=0
solve for y to get y = 4 and y = 16


since y = x^2 and you want to solve for x, now replace y with x^2 to get:


x^2 = 4 and x^2 = 16


solve for x to get x = plus or minus 2 and x = plus or minus 4


if you did this correctly, your graph crosses the x-axis at x = -2, +2, -4, +4


since your original equation is x^2-20x+64 >= 0, then you need to test each interval of the graph to see where the graph is greater than 0 and less than 0.


since your zero points are -2, +2, -4, +4, then the intervals you need to test are:


x < -4
-4 < x < -2
-2 < x < 2
2 < x < 4
x > 4


you test by picking a point in each interval and testing if the graph is positive or negative at that point.


for example, at the point x = -3, the graph of x^4 - 20x^2 + 64 is equal to (-3)^4 - 20(-3)^2 + 64 = -35 which is negative.


this means that the graph is negative in the interval -4 < x < -2


you do that at all the other intervals and you'll get a picture of when the graph is above the x-axis and below the x-axis.   


since the equation says that x^4 - 20x + 64 >= 0, then you pick the intervals that make that equation true.


as it turns out:


the intervals of x that satisfy the equation of x^2 - 20x + 64 >= 0 are:


x <= -4
-2 <= x <= 2
x >= 4


the graph of your equation is shown below:


<img src="http://theo.x10hosting.com/2014/092602.jpg" alt="$$$" </>


graphing is simpler, but if you didn't have graphing capability, then you needed to test each interval to find out if the equation was greater than or equal to 0 in that interval.