If the equations $ax^2+bx+c=0$ and $x^3+3x^2+3x+2=0$ have two common roots, then
- $a=b=c$
- $a=b\neq c$
- $a=-b=c$
-
none of these
The second equation x^3 + 3x^2 + 3x + 2 = 0 can be written as (x+1)^3 + 1 = 0. The roots are x+1 = -1, -omega, -omega^2. So x = -2, -1-omega, -1-omega^2. If the quadratic has two common roots, it must be a factor of the cubic. The only way to have two common roots is if the quadratic is (x+1+omega)(x+1+omega^2) = x^2 + 2x + 1 + omega + omega^2 + omega^3 = x^2 + 2x + 1 - 1 + 1 = x^2 + 2x + 1. Comparing with ax^2+bx+c=0, a=b=c.
The cubic equation x^3 + 3x^2 + 3x + 2 = 0 can be rewritten as (x + 1)^3 + 1 = 0. This gives (x + 2)((x + 1)^2 - (x + 1) + 1) = 0. The real root is x = -2 and the two complex roots are (1 +/- i sqrt(3))/2 - 1 = (-1 +/- i sqrt(3))/2. Since the quadratic equation shares these two complex roots it must be a non-zero multiple of (x - r1)(x - r2) = x^2 - (r1 + r2)x + r1 r2. The sum of the complex roots is -1 and their product is 1 so the quadratic is a multiple of x^2 + x + 1. This means the coefficients satisfy a = b = c. Therefore a = b = c.