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 $a:b:c=1:2:3$


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

  • $\begingroup$ For different points of view, see my answer. $\endgroup$
    – Jean Marie
    May 28 '21 at 13:57

Here is a direct answer to your question. You should have continued with a Gauss decomposition of the quadratic expression:



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

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.

  • 1
    $\begingroup$ Thanks for letting me know another approach to solve the problem, $\endgroup$ May 28 '21 at 15:55

Hint: Try dividing the entire equation at the end by $b^2$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.