Solve the following system of equations:
x + y + z = 6
x2 + y2 + z2 = 18
√x + √y + √z = 4
I believe the following represents a complete solution.
Let a,b,c = √x,√y,√z ; a,b,c >=0
eq1 a+b+c = 4
eq2 a^2+b^2+c^2 = 6
eq3 a^4+b^4+c^4 = 18
square eq1
a^2+b^2+c^2 + 2(ab+ac+bc) = 16
so, (ab+ac+bc) = 5 (eq4)
square eq2
a^4+b^4+c^4 + 2(a^2b^2+a^2c^2+b^2c^2) = 36
so, (a^2b^2+a^2c^2+b^2c^2) = 9 (eq5)
square eq4
(ab+ac+bc)^2 = 25
(a^2b^2+a^2c^2+b^2c^2) + 2abc(a+b+c) = 25
9 + 2abc(4) = 25
so, abc = 2 (eq6)
a+b+c = 4
abc = 2
ab+ac+bc = 5
a,b,c >=0
let me just check AM GM:
AM = (a+b+c)/3 = 4/3 or 1.333
GM = (a*b*c)^(1/3) = cube root of 2 about 1.26
AM > GM, so far so good
Subst c = 4-a-b
ab(4-a-b) = 2
4ab - ab(a+b) = 2 (eq7)
ab+a(4-a-b)+b(4-a-b) = 5
ab + 4(a+b) - (a+b)^2 = 5 (eq8)
Let p=ab and s=a+b
4p - ps = 2 (eq7')
p + 4s - s^2 = 5 (eq8')
s^2 - (p+4)s + 3(p+1) = 0
s = [(p+4) ± √(p^2 - 4p + 4)]/2
s = [(p+4) ± (p-2)]/2
s = {p+1, 3}
case 1: s = p+1
a+b = ab+1 True if at least one of a and/or b is 1.
wlog say a=1 and b can be anything
1+b+c = 4 b+c=3
(1)bc = 2 bc=2
b+c+bc = 5 bc=2 �" checks
we have a=1 and b+c=bc+1
so one of b and/or c must equal 1.
wlog say b=1, then c=2
{a,b,c} = {1,1,2} in any order
Therefore {x,y,z} = {1,1,4} in any order
case 2: if s=3, a+b=3, a+b+c=4, so c=1
if c=1:
ab(1) = 2
ab+a+b = ab+3 = 5, so ab=3 and ab=2
case 2 is rejected
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I modified Jer's Desmos by changing the first equation to:
z = 6 - x - y
https://www.desmos.com/calculator/7kgakgwaez
Edited on October 22, 2024, 4:12 pm
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Posted by Larry
on 2024-10-22 16:05:48 |
Solution
