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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 $$(c-b)(ab^2-ac^2)=(ac-ab)^2.$$

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    $\begingroup$ never thought of setting the first two equal to each other ? $\endgroup$ – user451844 Sep 2 '17 at 0:22
  • $\begingroup$ The common root of the given two equations is $a$. If $a\neq 0$, the other roots of the given equations are $c$ and $b$ respectively which obviously satisfy $x^2+ax+bc=0$. $\endgroup$ – Prasun Biswas Sep 2 '17 at 0:41
  • $\begingroup$ @PrasunBiswas, Could you please elaborate? I didn't get anything.. $\endgroup$ – pi-π Sep 2 '17 at 0:43
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    $\begingroup$ First, $a,b,c$ must be pairwise distinct and this should have been clearly stated (otherwise the claim is obviously false). Second, the condition $a\neq 0$ is missing. If $a=0$ and $b+c\neq 0$, then the hypothesis holds but the claim is false. $\endgroup$ – Batominovski Sep 2 '17 at 0:44
  • $\begingroup$ @ Roddy MacPhee, Setting them equal, I got either $b=c$, or $\alpha =a$?? $\endgroup$ – pi-π Sep 2 '17 at 0:46
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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.

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  • $\begingroup$ How is $b+c=-\beta_1 - \beta_2 -2a$? $\endgroup$ – pi-π Sep 2 '17 at 1:04
  • $\begingroup$ @blue_eyed_... Vieta's formula $\endgroup$ – Stefan4024 Sep 2 '17 at 1:23
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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.

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If $abc=0$,that's too trivial to need to discuss. We assume that $abc \neq 0$. Thus, all the roots do not equal $0.$

Let $\alpha, \beta $ be the roots for $x^2+bx+ca=0$. By Vieta's theorem, we have $$\alpha+\beta=-b,\tag1$$$$\alpha\beta=ca.\tag 2$$

Similarily,let $\alpha, \gamma $ be the roots for $x^2+cx+ab=0$. By Vieta's theorem, we have $$\alpha+\gamma=-c,\tag3$$$$\alpha\gamma=ab.\tag 4$$

Hence, by $(1)-(2)$, we have $$\beta-\gamma=-b+c.\tag5$$ By $(1) \div (2)$,we have $$\frac{\beta}{\gamma}=\frac{c}{b}. \tag6$$ From $(5)$ and $(6)$, we may obtain $$\beta=c,\tag7$$and$$\gamma=b.\tag8$$

Putting $(7)$ into $(2)$, we have $$\alpha=a.\tag9$$Putting $(7),(9)$ into $(1)$,we have $$a+c=-b.\tag{10}$$

Now, we may have $\beta+\gamma=c+b=c-(a+c)=-a,$and $\beta \gamma=bc.$ By Vieta's theorem, we may conclude that $\beta, \gamma$ are necessarily the roots for $x^2+a+bc=0.$ We are done!

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