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If the equations $x^2+2x+3=0$ and $ax^2+bx+c=0$; $\enspace$ $a,b,c \in \mathbb{R}$, have a common root, then $a:b:c$ is ?

This is my approach:
let the common root be $\alpha$. Then
$$\alpha^2+2\alpha+3=0------(1)$$ $$a\alpha^2+b\alpha+c=0------(2)$$.
Now the above two eqations can be solved in $\alpha \enspace and\enspace \alpha^2$.
$$\frac{\alpha^2}{2c-3b}=\frac{\alpha}{3a-c}=\frac{1}{b-2a}$$ $$\alpha^2=\frac{2c-3b}{b-2a};\alpha=\frac{3a-c}{b-2a}$$ $$\implies \frac{2c-3b}{b-2a}=\left(\frac{3a-c}{b-2a}\right)^2$$
On further simplifying we get: $$9a^2+3b^2+c^2-6ab-2ac-2bc=0$$.
After this I am unable to proceed further. I thought a lot on how to obtain their ratios, but I am stuck.
The ratio given is $a:b:c=1:2:3$

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Discriminant is negative. It means the roots are complex conjugate of each other. So, both equations have both roots common. So, $a:b:c=1:2:3$

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  • $\begingroup$ For different points of view, see my answer. $\endgroup$
    – Jean Marie
    May 28 '21 at 13:57
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Here is a direct answer to your question. You should have continued with a Gauss decomposition of the quadratic expression:

$$9a^2+3b^2+c^2-6ab-2ac-2bc=0\tag{1}$$

into:

$$(3a-b-c/3)^2+2(b-2c/3)^2=0 \ \ \iff \ \ \begin{cases}3a-b-c/3=0 & Eq. (2)\\b-2c/3=0& Eq. (3)\end{cases}$$

From (3), one deduces that $b=2c/3$ ; plugging this expression of $b$ into (2) gives $3a=c$. Therefore we have established the desired proportionality.

Remark 1: expression (1) can be directly obtained by calculating the so-called resultant

$$\det\begin{pmatrix}1&2&3&0\\ 0&1&2&3\\ a&b&c&0\\ 0&a&b&c \end{pmatrix}$$

Remark 2: There was a completely different way to solve this question. As the roots of the first polynomial are $-1+i\varepsilon \sqrt{3}$, plugging (one of) them into the second equation gives:

$$(-2a-b+c)+i\varepsilon \sqrt{3}(-2a+b)=0$$

Identifiying real part and imaginary part with zero gives at once the result.

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    $\begingroup$ Thanks for letting me know another approach to solve the problem, $\endgroup$ May 28 '21 at 15:55
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Hint: Try dividing the entire equation at the end by $b^2$

(You can divide by either of $c^2$ or $a^2$ though...)

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  • $\begingroup$ Thanks for you hint! $\endgroup$ May 28 '21 at 11:50

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