Question 1205913
subtract 20x^2 from both sides to get:
x^4 + 64 - 20x^2 = 0
arrange in descending order of degree to get:
x^4 - 20x^2 + 64 = 0
let y = x^2 to get:
y^2 - 20y + 64 = 0
factor to get:
(y-4) * (y-16) = 0
solve for y to get:
y = 4 or y = 16
since y = x^2, you get:
x = plus or minus 2 or x = plus or minus 4


your solutions are x = 2, -2, 4, -4


plug each of those into the original equation and the original equation will be true.


for example, when x = 4, x^4 + 64 = 80x^2 becomes 320 = 320, which is true and when x = 2, x^4 + 64 = 20x^2 becomes 80 = 80


here's what it looks like when you graph y = x^4 - 20x^2 + 64.


<img src = "http://theo.x10hosting.com/2024/020101.jpg">


here's what it looks like when you graph y = x^2 + 64 and you graph y = 20x^2.


<img src = "http://theo.x10hosting.com/2024/020102.jpg">


the first graph shows the intersection of the graphed equation with y = 0.


the second graph shows the intersection of the two graphed equations.


the x-value of the intersection is the same in both graphs.