Question 814491: There are numbers a, b, c, and d where the sum of any one of them and the product of the other three is equal to 2. If a>b, a>c, and a>d, find a.
Answer by Edwin McCravy(20054)   (Show Source):
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(1) a + bcd = 2
(2) b + acd = 2
(3) c + abd = 2
(4) d + abc = 2
Subtract eq (2) from eq (1)
a - b + bcd - acd = 0
a - b - acd + bcd = 0
(a-b) - cd(a-b) = 0
(a-b)(1-cd) = 0
Since a>b, a-b>0
so 1-cd = 0
-cd = -1
cd = 1
Subtract eq (3) from eq (1)
a - c + bcd - abd = 0
a - c - abd + bcd = 0
(a-c) - bd(a-c) = 0
(a-c)(1-bd) = 0
Since a>c, a-c>0
so 1-bd = 0
-bd = -1
bd = 1
Subtract eq (4) from eq (1)
a - d + bcd - abc = 0
a - d - abc + bcd = 0
(a-d) - bc(a-d) = 0
(a-d)(1-bc) = 0
Since a>d, a-d>0
so 1-bc = 0
-bc = -1
bc = 1
So we have
cd = 1
bd = 1
bc = 1
cd = bd = 1, so c = b
cd = bc = 1, so d = b
So b = c = d
a + bcd = 2
a + bbb = 2
a + b³ = 2
b + acd = 2
b + abb = 2
b + ab² = 2
So we have the system
a + b³ = 2
b + ab² = 2
Solve the 1st for a
a = 2 - b³
Substitute in the 2nd eq
b + (2-b³)b² = 2
b + 2b² - b5 = 2
-b5 + 2b² + b - 2 = 0
b5 - 2b² - b + 2 = 0
Possible rational roots are ±1,±2
1|1 0 0 -2 -1 2
| 1 1 1 -1 -2
1 1 1 -1 -2 0
(b-1)(b4+b³+b²-b-2)=0
1|1 1 1 -1 -2
| 1 2 3 2
1 2 3 2 0
(b-1)(b-1)(b³+2b²+3b+2)=0
-1|1 2 3 2
| -1 -1 -2
1 1 2 0
(b-1)(b-1)(b+1)(b²+b+2)=0
The last factor does not yield real
roots.
So the real roots are 1 and -1
b can't be 1 for
a + b³ = 2
a + 1³ = 2
a + 1 = 2
a = 1
which contradicts a>b
Therefore b=c=d=-1
a + b³ = 2
a + (-1)³ = 2
a - 1 = 2
a = 3
Edwin
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