This seems like it would take some calculations. If we use vietas formulas this thing looks very ugly... I will use n instead of √n in my calculations From first poly roots -20,m+ni,m-ni Coeff x^2:-20+2m=0 Coeff X: -20*2m+m^2-n^2=a Free coeff: -20(m^2+n^2)=-b -20^3-20a+b=0 -8000+800m-20m^2-20n^2+20(m^2+n^2)=0 -8000+800m=0 so m=10 From second polinomial -c=-21+2m=-1 => c=1 -d=-21(m^2+n^2) -21^3+21^2c+d=0 -21^2*20+d=0 d=20*21^2=21(m^2+n^2) 20*21=100+n^2 n^2=420-100=320 n=√320 According to original problem sum of Re+im^2=330
since the cubic polynomials have a single real root, the other two roots are complex conjugate, so they share both roots, so we can write the polynomials as (x+20)(x^2+ex+f) and (x+21)(x^2+ex+f), expanding this out and using the fact the quadratic term is zero in the first we get e=-20, and then using the fact that the linear term is zero in the second we get that f=420. Now a simple application of the quadratic formula gives us that m=20/2, n=420-100=320, so the sum is 330
Nice problem
This seems like it would take some calculations. If we use vietas formulas this thing looks very ugly... I will use n instead of √n in my calculations
From first poly roots -20,m+ni,m-ni
Coeff x^2:-20+2m=0
Coeff X: -20*2m+m^2-n^2=a
Free coeff: -20(m^2+n^2)=-b
-20^3-20a+b=0
-8000+800m-20m^2-20n^2+20(m^2+n^2)=0
-8000+800m=0 so m=10
From second polinomial
-c=-21+2m=-1 => c=1
-d=-21(m^2+n^2)
-21^3+21^2c+d=0
-21^2*20+d=0
d=20*21^2=21(m^2+n^2)
20*21=100+n^2
n^2=420-100=320
n=√320
According to original problem sum of Re+im^2=330
since the cubic polynomials have a single real root, the other two roots are complex conjugate, so they share both roots, so we can write the polynomials as (x+20)(x^2+ex+f) and (x+21)(x^2+ex+f), expanding this out and using the fact the quadratic term is zero in the first we get e=-20, and then using the fact that the linear term is zero in the second we get that f=420. Now a simple application of the quadratic formula gives us that m=20/2, n=420-100=320, so the sum is 330