If $\displaystyle x^2+abx+c=0 \mbox\ {and} \ x^2+acx+b=0$ quadratic equations have a common root.prove b=c or the other roots are the roots of $\displaystyle a(b+c)x^2+(b+c)-abc=0$
The roots of the first equation are
$\displaystyle \displaystyle \begin{align*} x &= \frac{-ab \pm \sqrt{(ab)^2 - 4(1)(c)}}{2(1)} \\ &= \frac{-ab \pm \sqrt{a^2b^2 - 4c}}{2} \end{align*}$
and the roots of the second equation are
$\displaystyle \displaystyle \begin{align*} x &= \frac{-ac \pm \sqrt{(ac)^2 - 4(1)(b)}}{2(1)} \\ &= \frac{-ac \pm \sqrt{a^2c^2 - 4b}}{2} \end{align*}$
Since they share a root, there is some value such that
$\displaystyle \displaystyle \begin{align*} \frac{-ab \pm \sqrt{a^2b^2 - 4c}}{2} &= \frac{-ac \pm \sqrt{a^2c^2 - 4b}}{2} \\ -ab \pm \sqrt{a^2b^2 - 4c} &= -ac \pm \sqrt{a^2c^2 - 4b} \end{align*}$
If $\displaystyle \displaystyle \begin{align*} b = c \end{align*}$, we have
$\displaystyle \displaystyle \begin{align*} LHS &= -ab \pm \sqrt{a^2b^2 - 4c} \\ &= -ab \pm \sqrt{a^2b^2 - 4b} \\ \\ RHS &= -ac \pm \sqrt{a^2c^2 - 4b} \\ &= -ab \pm \sqrt{a^2b^2 - 4b} \\ &= LHS \end{align*}$
So that means for the two equations to share a root, it's possible that $\displaystyle \displaystyle \begin{align*} b = c \end{align*}$.