Question

If $x^2+ax+bc=0$ and $x^2+bx+ca=0(a\ne b)$ have a common root, then prove that their other roots satisfy the equation $x^2+cx+ab=0$.

Answer 1

For quadratic equation $a{x}^2+bx+c=0$

Sum of roots $=-\frac{b}{a}$

Product of roots $=\frac{c}{a}$

Let $x^2+ax+bc=0$ have roots $\alpha$ and $\beta$:

$\alpha +\beta =-a......................\left(1\right)\alpha \beta =bc...................\left(2\right)$

Let $x^2+bx+ac=0$ have roots $\alpha$ and $\gamma$:

$\alpha +\gamma =-b..................\left(3\right)\alpha \gamma =ac..................\left(4\right)$

Subtracting (3) from (1) ,we get

$\beta -\gamma =b-a..................\left(5\right)$

Dividing (2) and (3)

$\frac{\beta }{\gamma }=\frac{b}{a}.................\left(6\right)$

From (5) and (6)

$\beta =b,\gamma =a$

From (2)

$\alpha .b=bc\Rightarrow \alpha =c$

Now, equation having other roots is

$x^2-(\beta +\gamma )x+\beta \gamma =0x^2-(a+b)x+ab..................\left(7\right)$

Since, $\beta =b$ is the root of $x^2+ax+bc=0$

$\Rightarrow b^2+ab+bc=0\Rightarrow a+b=-c$

Putting this in (7)

we get $x^2+cx+ab=0$