Equations $x^2+bx+ca=0$ and $x^2+cx+ab=0$ have a common root. Prove that equation $x^2+ax+bc=0$ contains two other roots

Question

Two quadratic equations are,

$$x^2+bx+ca=0$$ $$x^2+cx+ab=0$$

where $b$ and $c$ are not equal to zero, have one common root. Prove that the equation containing their other roots is

$$x^2+ax+bc=0$$

Answer 1

The hypotheses as stated are not quite right. The correct hypotheses are $b\ne c$ and $a\ne 0$.

Indeed, if $b=c$, then problem is ill posed. If $a=0$, then $x^2-x$ and $x^2-2x$ have $x=0$ as a common root. The other roots are $1$ and $2$, which are not roots of $x^2+2$.

So, let's assume that $b\ne c$ and $a\ne 0$.

Subtracting the two equations gives $0=(b-c)x+a(c-b)=(b-c)(x-a)$. Since $b\ne c$, we get that $a$ is the common root.

Looking at the independent terms in the two equations, and using that $a\ne0$, the other two roots are $b$ and $c$ and so are the roots of $x^2-(b+c)x+bc$. It remains to prove that $b+c=-a$.

Setting $x=a$ in $0=x^2+bx+ca$ gives $0=a^2+ab+ac=a(a+b+c)$ and so $b+c=-a$, since $a\ne0$.

Answer 2

Let $r$ and $r_1$ be the roots of $x^2+bx+ac=0$, and $r$ and $r_2$ be the roots of $x^2+cx+ab=0$.

$$rr_1=ac,\>\>\>\>\>rr_2=ab\tag{1}$$

$$r+r_1=-b, \>\>\>\>\>r+r_2=-c\tag{2}$$

which lead to,

$$ \frac{r_1}{r_2}=\frac cb,\>\>\>\>\>\frac{r_1}{r_2}=\frac{b+r}{c+r}=\frac cb\implies r=-(b+c)$$

$$2r+r_1+r_2=-(b+c)\implies r_1+r_2=b+c$$

$$a(b+c)=r(r_1+r_2)\implies a=r=-(b+c)$$

As a result,

$$r_1r_2 = \frac{a^2bc}{r^2}=bc$$ $$r_1+r_2=b+c=-a$$

Thus, $r_1$ and $r_2$ satisfy

$$x^2+ax + bc=0$$