Each of a and b is a real number that satisfies this system of equations:
a3+9a2*b=10
b3+a*b2=2
LHS1 = 5*LHS2
a^3 + 9a^2*b = 5b^3 + 5a*b^2
a^3 + 9a^2*b - 5a*b^2 - 5b^3 = 0
t = a/b
t^3 + 9t^2 - 5t - 5 = 0
sum of coefficients = 0, t=1 is a solution
t^3 - t^2 + 10t^2 - 10t + 5t - 5 = 0
(t-1)(t^2 + 10t + 5) = 0
t = {1, -5 ± 2√5) = a/b
Any solution must also solve original equations
case 1
t = 1 --> a=b=1
case 2
t = -5 + 2√5
a = (-5 + 2√5)b
b^3 + (-5 + 2√5)b^3 = 2
(-4 + 2√5)b^3 = 2
b^3 = 2 / (-4 + 2√5)
b^3 = 2(-4 - 2√5)/(-4)
b^3 = 2 + √5
find cube root to get b
(c+d√5)^3 = 2 + √5
c^3 + 3c^2d√5 + 3*5*cd^2 + 5d^3√5
c^3 + 3*5*cd^2 + (3c^2d + 5d^3)√5
c^3 + 15*cd^2 = 2
3c^2d + 5d^3 = 1
c^3 + 15*cd^2 = 2(3c^2d + 5d^3)
c^3 - 6c^2d + 15*cd^2 - 10d^3 = 0
u = c/d
u^3 - 6u^2 + 15*u - 10 = 0
again 1 is a solution
(u-1)(u^2 - 5u + 10); det<0 so u=1 only real solution
u=1, c=d 8c^3 = 1; c=d=1/2
We're still on case 2:
t = -5 + 2√5
b^3 = 2 + √5
b = (1 + √5)/2
a = (-5 + 2√5)b = (-5 + 2√5)(1 + √5)/2
a = (5 - 3√5)/2
case 3
t = -5 - 2√5
a = (-5 - 2√5)b
b^3 + (-5 - 2√5)b^3 = 2
(-4 - 2√5)b^3 = 2
b^3 = 2 / (-4 - 2√5)
b^3 = 2(-4 + 2√5)/(-4)
b^3 = 2 - √5 (note b<0)
not a surprise, conjugate of case 2
b = (1 - √5)/2
a = (5 + 3√5)/2
There are 3 solutions for (a,b):
(1,1) and
( (5 - 3√5)/2 , (1 + √5)/2 ) and
( (5 + 3√5)/2 , (1 - √5)/2 )
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Posted by Larry
on 2024-06-25 09:46:35 |
Solution