Home > Just Math
| Real Resolution (Posted on 2014-01-16) |
 |
Determine all possible real solutions to this system of equations:
A+√(B*C) = 29, B+√(C*A) = 31, C+√(A*B) = 37
Prove that there are no others.
|
No Solution Yet
|
Submitted by K Sengupta
|
|
Rating: 3.5000 (2 votes)
|
|
More solutions
|
Comment 2 of 2 | 
|
An integer solution has already been posted, but I think Ive found the other
solutions and would welcome any news of a shorter method.
Its clear that A, B, C must have the same sign for real square roots, so
substituting B = k2A is permissible and the three equations then become:
A + k*sqrt(AC) = 29; k2A + sqrt(AC) = 31; C + kA = 37
Eliminating sqrt(AC) between the first two gives A = (31k 29)/(k3 1),
thus B = k2(31k 29)/(k3 1) and, from the third equation, after some
rearrangement: C = (37k3 - 31k2 +29k 37)/(k3 1).
Substituting these results into the original equation, B + sqrt(CA) = 31,
after much algebra gives: 153k4 - 1017k3 +1798k2 994k + 56 = 0
which factorises to (3k 4)(51k3 -271k2 + 238k 14) = 0.
The linear factor yields k = 4/3 which gives the integer solution already
posted, while the cubic factor gives three irrational roots, two of which
give values of A, B and C that satisfy the original problem, as follows:
A B C
k = 4/3 9 16 25
k = 0.0633369.. 27.0434.. 0.108486.. 35.2872..
k = 4.22442.. 1.37061.. 24.4596.. 31.2099..
k = 1.02597.. not a valid solution.
|
|
Posted by Harry
on 2014-01-31 17:17:59 |
|
|
Please log in:
Forums (0)
Newest Problems
Random Problem
FAQ |
About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
|
