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Let $x_1, x_2$ be the roots of the equation $x^2 + ax + bc = 0$, and $x_2, x_3$ the roots of the equation $x^2 + bx + ac = 0$ with $ac \neq bc$. Show that $x_1, x_3$ are the roots of the equation $x^2 + cx + ab=0$.

From Vieta's I have: $\begin{cases} x_1+x_2=-a\\x_1x_2=bc\end{cases}$ $\begin{cases} x_2+x_3=-b\\x_2x_3=ac\end{cases}$

and I have to prove: $\begin{cases} x_1+x_3=-a\\x_1x_3=ab\end{cases}$

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    $\begingroup$ A common solution of the first two equations is a solution of their subtraction. Who should be $x_2$? Once you know $x_2$ you know the rest. $\endgroup$ – Pp.. Jan 23 '15 at 21:08
  • $\begingroup$ Why do you need to prove that $x_1+x_3 = -a$? If you do succeed, note that since you already know that $x_1+x_2=-a$ and have just proved that $x+1+x_3 = -a$, then it must be that $x_2 = x_3$, no? $\endgroup$ – Dilip Sarwate Jan 23 '15 at 21:10
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Since $ac\neq bc$ we have $a\neq b$ and $c\neq 0$.

It easy to see that $x_2$ is equal to $c$. In fact, $$ x_2^2+ax_2+bc=x_2^2+bx_2+ac $$ $$ (a-b)x_2=ac-bc $$ $$ x_2=c $$

From Vieta's we have $x_1=b$, $x_3=a$ and $b+c=-a$. The rest is obvious.

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Subtract the first two equations to get $x_2$

$$\begin{align}x_2^2+ax_2+bc&=0\\x_2^2+bx_2+ac&=0\end{align}$$ to get $$(a-b)x_2-(a-b)c=0$$ from where $x_2=c$. So, $$\begin{align}x_1&=b=-a-c\\x_3&=a=-b-c\end{align}$$ from where $$\begin{align}x_1+x_3&=a+b=c\\x_1x_3&=ab\end{align}$$

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