Prove that if the equations $x^2+bx+ca=0$ and $x^2+cx+ab=0$ have a common root,

Question

Prove that if the equations $x^2+bx+ca=0$ and $x^2+cx+ab=0$ have a common root, their other roots will satisfy $x^2+ax+bc=0$.

My Attempt:

Given, $$x^2+bx+ca=0$$ $$x^2+cx+ab=0$$ Let $\alpha $ be the common root of both our equations. So, $$\alpha^2 + b\alpha + ca=0$$ $$\alpha^2 + c\alpha +ab=0.$$ If one root is common, then $$(c-b)(ab^2-ac^2)=(ac-ab)^2.$$

Answer 1

As noted in the comments you will get that $a$ is the common root. Now you can use Vieta's formulas. Let $\beta_1$ and $\beta_2$ the other root of the first and second equation, respectively. Then we have:

$$a\beta_1 = ca$$ $$a\beta_2 = ab$$

So $\beta_1 = c$ and $\beta_2 = b$. Now the rest shouldn't be too hard. In fact you get by Vieta's forumlas that $b + c = -\beta_1 - \beta_2 - 2a \iff -a = b+c$. Then we get that the quadratic polynomial that have $b$ and $c$ as roots is:

$$(x-b)(x-c) = x^2 - (b + c)x + bc = x^2 + ax + bc$$

Hence the proof.

Answer 2

Your statement is wrong.

Try $b=c=6$ and $a=\frac{4}{3}$.

Given equations are $x^2+6x+8=0$, but the third equation $x^2+\frac{4}{3}x+36=0$ has no real roots.