If the equations $ax^2+bx+c=0$ and $x^3+3x^2+3x+2=0$ have two common roots then show that $a=b=c$.
My attempts:
Observing $-2$ is a root of $x^3+3x^2+3x+2=0\implies x^3+3x^2+3x+2=(x+2)(x^2+x+1)=0$
Hence $ax^2+bx+c=0$ can have complex roots in common, comming from $(x^2+x+1)=0$
Both the roots of $(x^2+x+1)=0$ and $ax^2+bx+c=0$ are common should imply $a=b=c$ not only this but $a=b=c=1$.
Is this solution correct?
We have that $$ \eqalign{ & x^{\,3} + 3x^{\,2} + 3x + 2 = 0\quad \Rightarrow \quad x^{\,3} + 3x^{\,2} + 3x + 1 = - 1\quad \Rightarrow \cr & \Rightarrow \quad \left( {x + 1} \right)^{\,3} = - 1\quad \Rightarrow \quad \left( {x + 1} \right) = e^{\,i\;\,\left( {1 + 2k} \right)\pi /3} = e^{\, \pm \,i\;\,\pi /3} ,e^{\,i\;\pi } = {1 \over 2} \pm i{{\sqrt 3 } \over 2},\;\; - 1 \cr} $$ and $$ \eqalign{ & 0 = ax^2 + bx + c\quad \Rightarrow \quad a\left( {x + 1} \right)^2 + \left( {b - 2a} \right)\left( {x + 1} \right) + \left( {a + c - b} \right) = 0\quad \Rightarrow \cr & \Rightarrow \quad \left( {x + 1} \right) = {{ - \left( {b - 2a} \right) \pm \sqrt {b^2 - 4ac} } \over {2a}} \cr} $$
Comparing the two results, the condition to have two identical roots imposes that it shall be $$ \left\{ \matrix{ {{2a - b} \over {2a}} = {1 \over 2} \hfill \cr {{\sqrt {4ac - b^2 } } \over {2a}} = {{\sqrt 3 } \over 2} \hfill \cr} \right.\;\quad \Rightarrow \quad \left\{ \matrix{ b = a \hfill \cr c = a \hfill \cr} \right. $$ either in the real and in the complex field.
Yes, I just check your calculus and are all correct. Your polynomial of grade $3$ defined as $x^3+3x^2+3x+2=0$ has one real root $x_{1}=-2$ and two complex roots $x_{2,3}=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$ so it can be written as $P(x)=(x+2)(x-(\frac{1}{2}-\frac{\sqrt{3}}{2}i))(x-(\frac{1}{2}+\frac{\sqrt{3}}{2}i))$. Then, remember the following: "a property of the second degree equations with real coefficients that have complex roots is that they are conjugated to each other". So, knowing this, there is no way you will have in common with the other polynomial one real root and other complex (is the only other option we have available), then: $a=b=c=1$.
If $a$, $b$, and $c$ are not real, then the question as stated is false. The roots of $x^3 + 3x^2 + 3x + 2$ are $-2$ and $-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$; and the polynomial $x^2 + (\frac{5}{2} + \frac{\sqrt{3}}{2} i)x + (1 + \sqrt{3} i) = (x + 2)(x + \frac{1}{2} + \frac{\sqrt{3}}{2} i)$ shares two roots with $x^3 + 3x^2 + 3x + 2$ but does not have $a = b = c$.
I think that the strongest statement that can be made is that if $a, b, c \in \mathbb{C}$ have the same argument, then the statement must be true. If this is the case, then we have $a = Ae^{i\theta}, b = B e^{i\theta}, c = e^{i \theta}$ for the same $\theta$ in each case and $A, B, C \in \mathbb{R}$. In such a case, the polynomial $a x^2 + bx + c = e^{i \theta}(A x^2 + Bx + C)$ will have the same roots as $Ax^2 + Bx + C$; and since this latter polynomial is real, its roots must be complex conjugates of each other. This implies that $A x^2 + Bx + C$ is a multiple of $(x + \frac{1}{2} + \frac{\sqrt{3}}{2} i)(x + \frac{1}{2} - \frac{\sqrt{3}}{2} i) = x^2 + x + 1$, and therefore that $A = B = C$. (Note, however, that they are not necessarily equal to 1.)
