The first equation has two roots:
±i
The second equation has two roots: (1/2)±(sqrt(3)/2)i
So there are actually four numbers that the problem's (i+ω) could represent.
(i) (1+(2+sqrt(3))i)/2
(ii) (1+(-2-sqrt(3))i)/2
(iii) (1+(2-sqrt(3))i)/2
(iv) (1+(-2+sqrt(3))i)/2
(i) and (ii) look like conjugate solutions to a quadratic equation ax^2+bx+c=0 where a=1, b=1 and the discriminant = -(2+sqrt(3))^2. Solving gives c=2+sqrt(3)
Polynomial g(x)=x^2+x+2+sqrt(3) then has the required zero (two of them) but has a non-integer coefficient.
(iii) and (iv) look like conjugate solutions to a quadratic equation ax^2+bx+c=0 where a=1, b=1 and the discriminant = -(2-sqrt(3))^2. Solving gives c=2-sqrt(3)
Polynomial h(x)=x^2+x+2-sqrt(3) then has the required zero (two of them) but has a non-integer coefficient.
The solution is to multiply them together. All the sqrt(3)'s cancel and we get a 4th degree polynomial that has all the zeros.
f(x)=g(x)*h(x)= 1x^4 + 2x^3 + 5x^2 + 4x + 1
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Posted by Jer
on 2021-11-23 09:53:35 |
Solution
