Question

If the equations $x^2+cx+ab=0$ and $x^2+bx+ca=0$ have a common root, show that $a+b+c=0$ also show that their other roots are and their other root are given by the equation $x^2+ax+bc=0$ (where $b\ne c$).

Answer 1

When two quadratics$x^2+Ax+B=0$ and $x^2+Cx+D=0$ have a common root, it’s $x=D-BA-C$.

Proof: Subtracting the quadratics yields the result.

So the common root of $x^2+bx+ac=0$ and $x^2+cx+ab=0$ is $x=ab-acb-c=a$.

Let’s call the other root of the first r and the other root of the second s.

$x^2+bx+ac=(x-r)(x-a)$ so equating coefficients $ar=ac$ or $r=c$ and $b=-c-a$ or $a=-(b+c)$

$x^2+ax+ab=(x-s)(x-a)$ $as=ab$ and $a=-s-a$. So $s=b$.

We have an equation with roots c and b, so it’s

$0=(x-c)(x-b)=x^2-(b+c)x+bc$

We note from above $a=-(b+c)$, and substituting yields

$0=x^2+ax+bc$