Let the roots be a,b,c,d, and suppose ab = cd.
abcd = 4 (product of roots)
-> (ab)^2 = (cd)^2 = 4
-> ab = cd = +- 2
a + b + c + d = 11/6 (sum of roots)
-> a + b + (ab)/d + (ab)/c = 11/6
-> acd + bcd + abc + abd = (11/6)*cd
-> (-11/3) = (11/6)*cd
-> cd = -2
-> ab = cd = -2 (discounting the possibility that ab = cd = 2)
now ab + ac + ad + bc + bd + cd = -13/3
-> 2*ab + ac + ad + bc + bd = -13/3
-> ac + ad + bc + bd = -1/3
-> (a+b)*(c+d) = -1/3 ...(A)
but remember (a+b) + (c+d) = 11/6 ... (B)
-> (a+b) - 1/(3*(a+b)) = 11/6 (can do the above since we know from (A) that (a + b) =/= 0)
-> (a+b)^2 - (11/6)*(a+b) - 1/3 = 0
-> 6*(a+b)^2 - 11*(a+b) - 2 = 0
-> (a+b) = (11 +- 13)/12 = -1/6 or 2
using a similar argument from (B),
-> (c+d) = -1/6 or 2
obviously (a + b) =/= (c + d) (if that was true, then a+b = c+d = -1/3 or 4 =/= 11/6)
so let's choose (a+b) = 2, (c+d) = -1/6 (we can do this arbitrarily, we could also chose in the other way round. It doesn't change the actual roots of the polynomial, it only changes which root we assign to a,b,c,d). Then we have:
a + b = 2 AND ab = -2 ...(C)
-> a - 2/a = 2
-> a^2 - 2a - 2 = 0
-> a = (2 +- sqrt(12))/2 = 1 +- sqrt(3)
and similarly b = 1 +- sqrt(3)
but it's obvious that a =/= b from part (C)
so we can arbitrarily choose a = 1 + sqrt(3) and b = 1 - sqrt(3) (choosing a and b the other way round is perfectly acceptable too)
c + d = -1/6 and cd = -2 ...(D)
-> c - 2/c = -1/6
-> c^2 + c/6 - 2 = 0
-> 6c^2 + c - 12 = 0
-> c = (-1 +- 17)/12 = 4/3 or -3/2.
so like before you can choose c = 4/3 and d = -3/2 or vice versa.
So the roots are 1 + sqrt(3), 1 - sqrt(3), 4/3, and -3/2. There's probably a better way to do it I can't think of.
CrashOveride said:
I get:
1 + sqrt[3]
1 - sqrt[3]
1+1/3
-1.5
How did you get the solution so quick?
