Question

If one root of the equation $x^2+ax+b=0$ is a root of the equation $x^2+cx+d=0$, prove that the other roots satisfy the equation $x^2+x(2a-c)+(a^2-ac+d)=0$.

Answer 1

Both the equations have coefficient of unity and hence common root is obtained by subtracting the two equations.
$\therefore$ $x(a-c)+(b-d)=0$
$\therefore$ $\alpha =-\frac{(b-d)}{(a-c)}$
If the other other root of first equation be $\beta$, then
$\alpha +\beta =-a$ $\therefore$ $\beta =-a-\alpha$
or $\beta =-a+\frac{b-d}{a-c}=\frac{-a^2+ac+b-d}{a-c}$
$\therefore$ $\beta (a-c)+a^2-ac-b+d=0$..........(1)
Since $\beta$ is a root of (1)st
$\therefore$ ${\beta }^2+\alpha \beta +b=0$........(2)
Add the above two, we get the results (1) and (2).