I first created a system of equations rigorously (sorry my formatting is off):
(a + b(root2)) x (c + d(root2)) = -6 + 2(root2)
(a + b(root2)) x (e + f(root2)) = 2 - 4(root2)
(c + d(root2)) x (e + f(root2)) = 7 + 7(root2)
To get a set of 6 equations for 6 unknown variables:
ac + 2bd = -6
bc + ad = 2
ae + 2bf = 2
be + af = -4
ce + 2df = 7
de + cf = 7
My matrix background is shaky at best, so while I know creating a matrix would net a unique solution, I was certain I'd make a math error along the way, so I tried instead to deduce some things logically.
First I noted that from equation 5, both c and e must be odd.
Then, from equation 6, either d is odd and f is even, or vice versa.
Then I took a leap of faith that for all these sums to be bounded by -6 and 7, all 6 values should be fairly small. And since equations 5 and 6 are both positive sums, all coefficients are likely positive as well.
From that, I tried a couple guesses, but then I hit paydirt with c=1, d=2, e=3, f=1.
Once I managed success from that combination, I tested it back into equations 1 & 2.
This yielded 2 equations with only 2 unknown variables, a much more manageable task.
Equation 1 became a + 4b = -6
and Equation 2 became 2a + b = 2
resulting in a = 2 and b = -2.
I checked that result into equations 3 & 4, and everything checked on the math, but to make sure I didn't violate any rules of algebra, I checked the arithmagon and the math worked out nicely.
Thanks for reviewing my solution. I did go back and follow your steps of cross-multiplying and taking the square root, but still had to make some intuitive guesses as to my values. So I think in either case their is a little guess work involved, but since my solution took a very different path I thought I'd share it with you.
Love your challenges.
Ed Groth
Wyoming Valley West High School