Determine all possible real solutions to this system of equations:
x+y+z=30
x
2+y
2+z
2=300
I found one real and one complex:
x + y + z = 30
x^2 + y^2 + z^2 = 300
(x+y+z)^2 = 900
(x+y+z)^2 = (x^2 + y^2 + z^2) + 2(xy+xz+yz)
xy+xz+yz = 300
xy + (x+y)z = 300
z = 30 - (x+y)
Let s = x+y
Let p = x*y
p + (s)(30 - s) = 300
s^2 - 30s + (300-p) = 0
s = [30 ± √(4p - 300)]/2
s = 15 ± √(p-75)
if p = 100, s = {10,20}
case: s=20
x+y = 20
xy = 100
{x,y} = {10,10} --> {x,y,z} = {10, 10, 10}
case: s=10
x+y = 10
xy = 100
y = 100/x
x + 100/x = 10
x^2 - 10x + 100 = 0
x = 5(1 ± √3 i)
{x,y} = {5+5√3i,5-5√3i} --> {x,y,z} = {5+5√3i, 5-5√3i, 20}
|
|
Posted by Larry
on 2024-09-03 10:31:11 |
One real, but ...
